P1.T2. Quantitative Analysis

Practice questions for Quantitative Analysis: Econometrics, MCS, Volatility, Probability Distributions and VaR (Intro)

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  1. Suzanne Evans

    P1.T2.300. Probability functions (Miller)

    x = (125*p)^(1/3) x = [125^(1/3) ]*[ p^(1/3)] = 5*p^(1/3) because 125^(1/3 )= 5 another example can be 4*(1/2) = 2 this can also be solved like : (2*2)^(1/2) where 4 = 2*2 we can perform the square root ^(1/2) onto each so we get [2^(1/2) ]* [2^(1/2)] = root2 * root2 = 2
    x = (125*p)^(1/3) x = [125^(1/3) ]*[ p^(1/3)] = 5*p^(1/3) because 125^(1/3 )= 5 another example can be 4*(1/2) = 2 this can also be solved like : (2*2)^(1/2) where 4 = 2*2 we can perform the square root ^(1/2) onto each so we get [2^(1/2) ]* [2^(1/2)] = root2 * root2 = 2
    x = (125*p)^(1/3) x = [125^(1/3) ]*[ p^(1/3)] = 5*p^(1/3) because 125^(1/3 )= 5 another example can be 4*(1/2) = 2 this can also be solved like : (2*2)^(1/2) where 4 = 2*2 we can perform the square root ^(1/2) onto each so we get [2^(1/2) ]* [2^(1/2)] = root2 * root2 = 2
    x = (125*p)^(1/3) x = [125^(1/3) ]*[ p^(1/3)] = 5*p^(1/3) because 125^(1/3 )= 5 another example can be 4*(1/2) = 2 this can also be solved like : (2*2)^(1/2) where 4 = 2*2 we can perform the...
    Replies:
    59
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    1,756
  2. asocialnot

    Question 202.2: Variance of sum of random variables

    Hi asocialnot, Great question. Because 202.2 is looking for the variance of the sum of three random variables, each with its own distributional parameters. Your formula above, indeed works, but for each of the random variables itself. For example, the first bond has PD = 4% and, as it is a Bernoulli, we know the variance = 96%*4% = 3.840%. Consistent with the worked solution, then: Variance...
    Hi asocialnot, Great question. Because 202.2 is looking for the variance of the sum of three random variables, each with its own distributional parameters. Your formula above, indeed works, but for each of the random variables itself. For example, the first bond has PD = 4% and, as it is a Bernoulli, we know the variance = 96%*4% = 3.840%. Consistent with the worked solution, then: Variance...
    Hi asocialnot, Great question. Because 202.2 is looking for the variance of the sum of three random variables, each with its own distributional parameters. Your formula above, indeed works, but for each of the random variables itself. For example, the first bond has PD = 4% and, as it is a...
    Hi asocialnot, Great question. Because 202.2 is looking for the variance of the sum of three random variables, each with its own distributional parameters. Your formula above, indeed works, but...
    Replies:
    1
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    1,028
  3. David Harper CFA FRM

    P1.T2.202. Variance of sum of random variables

    David, please ignore me. I figured it out - the beta is between G and the portfolio and not G with S so I worked out part of the covariance but not the full covariance. I now understand the formula you have used. Sorry for the trouble.
    David, please ignore me. I figured it out - the beta is between G and the portfolio and not G with S so I worked out part of the covariance but not the full covariance. I now understand the formula you have used. Sorry for the trouble.
    David, please ignore me. I figured it out - the beta is between G and the portfolio and not G with S so I worked out part of the covariance but not the full covariance. I now understand the formula you have used. Sorry for the trouble.
    David, please ignore me. I figured it out - the beta is between G and the portfolio and not G with S so I worked out part of the covariance but not the full covariance. I now understand the...
    Replies:
    53
    Views:
    1,010
  4. Pam Gordon

    P1.T2.309. Probability Distributions I, Miller Chapter 4

    Hi @mojgan_sepehr it is like 100 2nd nCr 5*(0.05^5)*(0.95^(100-5)=0.1800 or 18%. Hope that helps! Thank you:)!
    Hi @mojgan_sepehr it is like 100 2nd nCr 5*(0.05^5)*(0.95^(100-5)=0.1800 or 18%. Hope that helps! Thank you:)!
    Hi @mojgan_sepehr it is like 100 2nd nCr 5*(0.05^5)*(0.95^(100-5)=0.1800 or 18%. Hope that helps! Thank you:)!
    Hi @mojgan_sepehr it is like 100 2nd nCr 5*(0.05^5)*(0.95^(100-5)=0.1800 or 18%. Hope that helps! Thank you:)!
    Replies:
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    997
  5. Pam Gordon

    P1.T2.310. Probability Distributions II, Miller Chapter 4

    Hi Dave, Maybe but you have triggered me into fixing my TI settings to AOS which is awesome. Feel much more comfortable with it. It was driving me mad on the chain setting and I could not figure out what was going on. Maybe we need a forum topic with the calculator settings and shortcuts in one place. This kind of knowledge is gold dust. Thanks Brendan
    Hi Dave, Maybe but you have triggered me into fixing my TI settings to AOS which is awesome. Feel much more comfortable with it. It was driving me mad on the chain setting and I could not figure out what was going on. Maybe we need a forum topic with the calculator settings and shortcuts in one place. This kind of knowledge is gold dust. Thanks Brendan
    Hi Dave, Maybe but you have triggered me into fixing my TI settings to AOS which is awesome. Feel much more comfortable with it. It was driving me mad on the chain setting and I could not figure out what was going on. Maybe we need a forum topic with the calculator settings and shortcuts in one...
    Hi Dave, Maybe but you have triggered me into fixing my TI settings to AOS which is awesome. Feel much more comfortable with it. It was driving me mad on the chain setting and I could not figure...
    Replies:
    41
    Views:
    849
  6. Suzanne Evans

    P1.T2.209 T-statistic and confidence interval

    Hi @FRMeugene Yes, I totally understand. @Nicole Manley started a thread here (which will eventually show up in our new Resources section) I hope this is helpful, thanks!
    Hi @FRMeugene Yes, I totally understand. @Nicole Manley started a thread here (which will eventually show up in our new Resources section) I hope this is helpful, thanks!
    Hi @FRMeugene Yes, I totally understand. @Nicole Manley started a thread here (which will eventually show up in our new Resources section) I hope this is helpful, thanks!
    Hi @FRMeugene Yes, I totally understand. @Nicole Manley started a thread here (which will eventually show up in our new Resources section) I hope this is helpful, thanks!
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    827
  7. Nicole Seaman

    P1.T2.312. Mixture distributions

    Just to add a few more thoughts, the exam "could" ask you to use an obscure level of significance which would require you to retrieve a value from a z table. If this was the case, the exam would provide a snippet of the respective region of the z table. (I would add that this is a totally reasonable question in my mind). Also, memorizing the most common z's will help you but I don't think...
    Just to add a few more thoughts, the exam "could" ask you to use an obscure level of significance which would require you to retrieve a value from a z table. If this was the case, the exam would provide a snippet of the respective region of the z table. (I would add that this is a totally reasonable question in my mind). Also, memorizing the most common z's will help you but I don't think...
    Just to add a few more thoughts, the exam "could" ask you to use an obscure level of significance which would require you to retrieve a value from a z table. If this was the case, the exam would provide a snippet of the respective region of the z table. (I would add that this is a totally...
    Just to add a few more thoughts, the exam "could" ask you to use an obscure level of significance which would require you to retrieve a value from a z table. If this was the case, the exam would...
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    43
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    815
  8. chris.leupold@baml.com

    question on: 208.3.C and 202.5

    Hi Chris, I think you are correct on both, can you see the source question thread @ i.e., you've identified two errors. I apologize they are not yet fixed in the PDF (like all errors, we will revise the PDFs, but I felt it more helpful currently to prioritize the 2 fresh mock exams). Thanks,
    Hi Chris, I think you are correct on both, can you see the source question thread @ i.e., you've identified two errors. I apologize they are not yet fixed in the PDF (like all errors, we will revise the PDFs, but I felt it more helpful currently to prioritize the 2 fresh mock exams). Thanks,
    Hi Chris, I think you are correct on both, can you see the source question thread @ i.e., you've identified two errors. I apologize they are not yet fixed in the PDF (like all errors, we will revise the PDFs, but I felt it more helpful currently to prioritize the 2 fresh mock exams). Thanks,
    Hi Chris, I think you are correct on both, can you see the source question thread @ i.e., you've identified two errors. I apologize they are not yet fixed in the PDF (like all errors, we will...
    Replies:
    11
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    792
  9. Nicole Seaman

    P1.T2.504. Copulas (Hull)

    Hello The practice questions that David writes are focused around the learning objectives in the GARP curriculum, but many times, his questions are more difficult. He writes them at a higher level to ensure that our members understand the concepts in depth. So while this question may be more difficult than the questions that you will see on the exam, the concepts are still testable, as they...
    Hello The practice questions that David writes are focused around the learning objectives in the GARP curriculum, but many times, his questions are more difficult. He writes them at a higher level to ensure that our members understand the concepts in depth. So while this question may be more difficult than the questions that you will see on the exam, the concepts are still testable, as they...
    Hello The practice questions that David writes are focused around the learning objectives in the GARP curriculum, but many times, his questions are more difficult. He writes them at a higher level to ensure that our members understand the concepts in depth. So while this question may be more...
    Hello The practice questions that David writes are focused around the learning objectives in the GARP curriculum, but many times, his questions are more difficult. He writes them at a higher...
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    30
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    745
  10. David Harper CFA FRM

    L1.T2.111 Binomial & Poisson

    Hi @bpdulog the numerator is (500*499*498*497*496) = 255,244,687,600; i.e., ~0.0004 * missing 499 ~= correct value. Re: 111.4, exp(-5) is e^(-5), see
    Hi @bpdulog the numerator is (500*499*498*497*496) = 255,244,687,600; i.e., ~0.0004 * missing 499 ~= correct value. Re: 111.4, exp(-5) is e^(-5), see
    Hi @bpdulog the numerator is (500*499*498*497*496) = 255,244,687,600; i.e., ~0.0004 * missing 499 ~= correct value. Re: 111.4, exp(-5) is e^(-5), see
    Hi @bpdulog the numerator is (500*499*498*497*496) = 255,244,687,600; i.e., ~0.0004 * missing 499 ~= correct value. Re: 111.4, exp(-5) is e^(-5), see
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    718
  11. Nicole Seaman

    P1.T2.503. One-factor model (Hull)

    @hellohi, This is how I have solved: e1=z1= -0.88 e2= pz1 + z2*sqrt(1-p^2) e2= [0.70*(-0.88)] + [0.63*sqrt(1-(0.7)^2) e2= -0.16609 U= Mean + (SD*e1) U= 5 + [3*(-0.88)] U= 2.36 V= Mean + (SD*e2) V= 10 + [6*(-0.16609)] V= 9.00346 Thanks, Rajiv
    @hellohi, This is how I have solved: e1=z1= -0.88 e2= pz1 + z2*sqrt(1-p^2) e2= [0.70*(-0.88)] + [0.63*sqrt(1-(0.7)^2) e2= -0.16609 U= Mean + (SD*e1) U= 5 + [3*(-0.88)] U= 2.36 V= Mean + (SD*e2) V= 10 + [6*(-0.16609)] V= 9.00346 Thanks, Rajiv
    @hellohi, This is how I have solved: e1=z1= -0.88 e2= pz1 + z2*sqrt(1-p^2) e2= [0.70*(-0.88)] + [0.63*sqrt(1-(0.7)^2) e2= -0.16609 U= Mean + (SD*e1) U= 5 + [3*(-0.88)] U= 2.36 V= Mean + (SD*e2) V= 10 + [6*(-0.16609)] V= 9.00346 Thanks, Rajiv
    @hellohi, This is how I have solved: e1=z1= -0.88 e2= pz1 + z2*sqrt(1-p^2) e2= [0.70*(-0.88)] + [0.63*sqrt(1-(0.7)^2) e2= -0.16609 U= Mean + (SD*e1) U= 5 + [3*(-0.88)] U= 2.36 V= Mean +...
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    20
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    711
  12. Fran

    P1.T2.301. Miller's probability matrix

    For working out mean of f(x), we integrate xf(x) instead of just integrating f(x) like the green statement above. Integrating xf(x) is just integrating x*f(x), i.e. you have another x, so not tricky to do as you know how to solve the green statement above after integrating xf(x), you can solve it by putting x = 6.
    For working out mean of f(x), we integrate xf(x) instead of just integrating f(x) like the green statement above. Integrating xf(x) is just integrating x*f(x), i.e. you have another x, so not tricky to do as you know how to solve the green statement above after integrating xf(x), you can solve it by putting x = 6.
    For working out mean of f(x), we integrate xf(x) instead of just integrating f(x) like the green statement above. Integrating xf(x) is just integrating x*f(x), i.e. you have another x, so not tricky to do as you know how to solve the green statement above after integrating xf(x), you can solve...
    For working out mean of f(x), we integrate xf(x) instead of just integrating f(x) like the green statement above. Integrating xf(x) is just integrating x*f(x), i.e. you have another x, so not...
    Replies:
    23
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    709
  13. Suzanne Evans

    P1.T2.303 Mean and variance of continuous probability density functions (pdf)

    Hi @hellohi fwiw, for the upcoming exam, I would not worry about this calculus. But to answer your question, the first expression is the variance of a continuous random variable (ie, Miller 3.21). We can solve for the mean, µ = 4, such that the integral contains the expression (x - µ)^2 * f(x) = (x - 4)^2 * f(x). We can then expand (x - 4)^2 = x^2 - 8x + 16 because (x -y)^2 = x^2 -2*x*y - y^2,...
    Hi @hellohi fwiw, for the upcoming exam, I would not worry about this calculus. But to answer your question, the first expression is the variance of a continuous random variable (ie, Miller 3.21). We can solve for the mean, µ = 4, such that the integral contains the expression (x - µ)^2 * f(x) = (x - 4)^2 * f(x). We can then expand (x - 4)^2 = x^2 - 8x + 16 because (x -y)^2 = x^2 -2*x*y - y^2,...
    Hi @hellohi fwiw, for the upcoming exam, I would not worry about this calculus. But to answer your question, the first expression is the variance of a continuous random variable (ie, Miller 3.21). We can solve for the mean, µ = 4, such that the integral contains the expression (x - µ)^2 * f(x) =...
    Hi @hellohi fwiw, for the upcoming exam, I would not worry about this calculus. But to answer your question, the first expression is the variance of a continuous random variable (ie, Miller 3.21)....
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  14. Fran

    P1.T2.307. Skew and Kurtosis (Miller)

    3rd central moment = ((1-0.05)^3)*5% + ((0-0.05)^3)*95% = 4.2750% 4th central moment = ((1-0.05)^4)*5% + ((0-0.05)^4)*95% = 4.0731% Note that these are central moments and NOT standardized central moments - the standardized central moments are divided by sigma ^ n where n = 3 for skewness (=3rd central moment) and n = 4 for kurtosis (=4th central moment)
    3rd central moment = ((1-0.05)^3)*5% + ((0-0.05)^3)*95% = 4.2750% 4th central moment = ((1-0.05)^4)*5% + ((0-0.05)^4)*95% = 4.0731% Note that these are central moments and NOT standardized central moments - the standardized central moments are divided by sigma ^ n where n = 3 for skewness (=3rd central moment) and n = 4 for kurtosis (=4th central moment)
    3rd central moment = ((1-0.05)^3)*5% + ((0-0.05)^3)*95% = 4.2750% 4th central moment = ((1-0.05)^4)*5% + ((0-0.05)^4)*95% = 4.0731% Note that these are central moments and NOT standardized central moments - the standardized central moments are divided by sigma ^ n where n = 3 for skewness (=3rd...
    3rd central moment = ((1-0.05)^3)*5% + ((0-0.05)^3)*95% = 4.2750% 4th central moment = ((1-0.05)^4)*5% + ((0-0.05)^4)*95% = 4.0731% Note that these are central moments and NOT standardized...
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    615
  15. Suzanne Evans

    P1.T2.212. Difference between two means

    That was a long message to type on a phone - got kind of tired towards the end!
    That was a long message to type on a phone - got kind of tired towards the end!
    That was a long message to type on a phone - got kind of tired towards the end!
    That was a long message to type on a phone - got kind of tired towards the end!
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    613
  16. Suzanne Evans

    P1.T2.206. Variance of sample average

    I am asking kind of dumb question, but where is this formula in the Miller Chapter (please tell me reference in David's Pdf)
    I am asking kind of dumb question, but where is this formula in the Miller Chapter (please tell me reference in David's Pdf)
    I am asking kind of dumb question, but where is this formula in the Miller Chapter (please tell me reference in David's Pdf)
    I am asking kind of dumb question, but where is this formula in the Miller Chapter (please tell me reference in David's Pdf)
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  17. LL

    209.1

    Thanks David ! This helps ! :) I went crazy figuring out how 27.8% is derived. Glad I asked :) You get a smile looking at my avatar. :D I hope I get a smile looking at my FRM result ! ;) :rolleyes:
    Thanks David ! This helps ! :) I went crazy figuring out how 27.8% is derived. Glad I asked :) You get a smile looking at my avatar. :D I hope I get a smile looking at my FRM result ! ;) :rolleyes:
    Thanks David ! This helps ! :) I went crazy figuring out how 27.8% is derived. Glad I asked :) You get a smile looking at my avatar. :D I hope I get a smile looking at my FRM result ! ;) :rolleyes:
    Thanks David ! This helps ! :) I went crazy figuring out how 27.8% is derived. Glad I asked :) You get a smile looking at my avatar. :D I hope I get a smile looking at my FRM result ! ;) :rolleyes:
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  18. David Harper CFA FRM

    L1.T2.104 Exponentially weighted moving average (EWMA)

    @Deepak Chitnis and @David Harper CFA FRM CIPM thanks for your replies...I will make sure I keep a special eye out as to whether the question mentions simple vs LN returns. If the question mentions neither, I think I shall plumb for the LN option as that just feels more "right" to me. But hopefully it won't be too much of an issue.
    @Deepak Chitnis and @David Harper CFA FRM CIPM thanks for your replies...I will make sure I keep a special eye out as to whether the question mentions simple vs LN returns. If the question mentions neither, I think I shall plumb for the LN option as that just feels more "right" to me. But hopefully it won't be too much of an issue.
    @Deepak Chitnis and @David Harper CFA FRM CIPM thanks for your replies...I will make sure I keep a special eye out as to whether the question mentions simple vs LN returns. If the question mentions neither, I think I shall plumb for the LN option as that just feels more "right" to me. But...
    @Deepak Chitnis and @David Harper CFA FRM CIPM thanks for your replies...I will make sure I keep a special eye out as to whether the question mentions simple vs LN returns. If the question...
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    526
  19. Fran

    P1.T2.304. Covariance (Miller)

    Hi @bpdulog You are referring to Covariance Method 1 (see above in the answer). The E[X] = -3.0%*30% + 1.0%*50% + 5.0%*20% = -0.900% + 0.500% + 1.000% [see fourth column above, under Exp Value X] = 0.600% is the expected value of X because it's just a weighted average. Then see under Covariance Method 1, the first row is given by (-3.0% - 0.6000%)*(-2.0% - 1.000%)*30% = 0.0324%; i.e., (Xi -...
    Hi @bpdulog You are referring to Covariance Method 1 (see above in the answer). The E[X] = -3.0%*30% + 1.0%*50% + 5.0%*20% = -0.900% + 0.500% + 1.000% [see fourth column above, under Exp Value X] = 0.600% is the expected value of X because it's just a weighted average. Then see under Covariance Method 1, the first row is given by (-3.0% - 0.6000%)*(-2.0% - 1.000%)*30% = 0.0324%; i.e., (Xi -...
    Hi @bpdulog You are referring to Covariance Method 1 (see above in the answer). The E[X] = -3.0%*30% + 1.0%*50% + 5.0%*20% = -0.900% + 0.500% + 1.000% [see fourth column above, under Exp Value X] = 0.600% is the expected value of X because it's just a weighted average. Then see under Covariance...
    Hi @bpdulog You are referring to Covariance Method 1 (see above in the answer). The E[X] = -3.0%*30% + 1.0%*50% + 5.0%*20% = -0.900% + 0.500% + 1.000% [see fourth column above, under Exp Value X]...
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  20. LL

    63.1

    Thanks David!
    Thanks David!
    Thanks David!
    Thanks David!
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  21. Fran

    P1.T2.305. Minimum variance hedge (Miller)

    Hi @irwinchung I finally figured it out but i had to model it out in additional pages in the CAPM learning sheet, before I could see it. Feel free to take a look at sheets "BT p1-t2-305-1" versus "BT p1-t2-305-2" in this workbook The difference is that in the hedge scenario (305.2) we assume asset A = $100 (as Miller assumes $1.00) such that various hedge weights assigned to Asset B are...
    Hi @irwinchung I finally figured it out but i had to model it out in additional pages in the CAPM learning sheet, before I could see it. Feel free to take a look at sheets "BT p1-t2-305-1" versus "BT p1-t2-305-2" in this workbook The difference is that in the hedge scenario (305.2) we assume asset A = $100 (as Miller assumes $1.00) such that various hedge weights assigned to Asset B are...
    Hi @irwinchung I finally figured it out but i had to model it out in additional pages in the CAPM learning sheet, before I could see it. Feel free to take a look at sheets "BT p1-t2-305-1" versus "BT p1-t2-305-2" in this workbook The difference is that in the hedge scenario (305.2) we assume...
    Hi @irwinchung I finally figured it out but i had to model it out in additional pages in the CAPM learning sheet, before I could see it. Feel free to take a look at sheets "BT p1-t2-305-1" versus...
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    12
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    474
  22. Nicole Seaman

    P1.T2.502. Covariance updates with EWMA and GARCH(1,1) models

    that helps much .....thanks a lot dear deepak....;)
    that helps much .....thanks a lot dear deepak....;)
    that helps much .....thanks a lot dear deepak....;)
    that helps much .....thanks a lot dear deepak....;)
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  23. Fran

    P1.T2.306. Calculate the mean and variance of sums of variables.

    Hi @hellohi Re 306.2 a^2 cancels in (a^2 + 1 - a^2). Imagine a^2 = y, then we have (y + 1 - y) = 1.0. The square, a^2, is evaluated before + or - In the expression, 1.0 + 2*cov[a*F, sqrt(1-a^2)*e], it is important to distinguish between variables and constants because (a,b) are the constants and (x,y) are the variables in the key covariance property which allows us to "take the constants out"...
    Hi @hellohi Re 306.2 a^2 cancels in (a^2 + 1 - a^2). Imagine a^2 = y, then we have (y + 1 - y) = 1.0. The square, a^2, is evaluated before + or - In the expression, 1.0 + 2*cov[a*F, sqrt(1-a^2)*e], it is important to distinguish between variables and constants because (a,b) are the constants and (x,y) are the variables in the key covariance property which allows us to "take the constants out"...
    Hi @hellohi Re 306.2 a^2 cancels in (a^2 + 1 - a^2). Imagine a^2 = y, then we have (y + 1 - y) = 1.0. The square, a^2, is evaluated before + or - In the expression, 1.0 + 2*cov[a*F, sqrt(1-a^2)*e], it is important to distinguish between variables and constants because (a,b) are the constants...
    Hi @hellohi Re 306.2 a^2 cancels in (a^2 + 1 - a^2). Imagine a^2 = y, then we have (y + 1 - y) = 1.0. The square, a^2, is evaluated before + or - In the expression, 1.0 + 2*cov[a*F,...
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  24. David Harper CFA FRM

    L1.T2.108 Volatility forecast with GARCH(1,1)

    Hi @Tania Pereira Right, either is acceptable and, in the case of question 108.3 above, it makes a difference: the given answer is 2.363% but if we instead computed a discrete daily return (i.e., 11.052/10 - 1 = 3.83%) then the 10-day volatility forecast is 2.429%, a difference of 0.066%. That's why this older question of mine is clearly imprecise (sorry): the question needs to specify that...
    Hi @Tania Pereira Right, either is acceptable and, in the case of question 108.3 above, it makes a difference: the given answer is 2.363% but if we instead computed a discrete daily return (i.e., 11.052/10 - 1 = 3.83%) then the 10-day volatility forecast is 2.429%, a difference of 0.066%. That's why this older question of mine is clearly imprecise (sorry): the question needs to specify that...
    Hi @Tania Pereira Right, either is acceptable and, in the case of question 108.3 above, it makes a difference: the given answer is 2.363% but if we instead computed a discrete daily return (i.e., 11.052/10 - 1 = 3.83%) then the 10-day volatility forecast is 2.429%, a difference of 0.066%. That's...
    Hi @Tania Pereira Right, either is acceptable and, in the case of question 108.3 above, it makes a difference: the given answer is 2.363% but if we instead computed a discrete daily return (i.e.,...
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    391
  25. Suzanne Evans

    P1.T2.208. Sample mean estimators (Stock & Watson)

    Hi David, I was just referring to the previous discussion to give better understanding to my question:) Thanks a lot for your time and patience. Praveen
    Hi David, I was just referring to the previous discussion to give better understanding to my question:) Thanks a lot for your time and patience. Praveen
    Hi David, I was just referring to the previous discussion to give better understanding to my question:) Thanks a lot for your time and patience. Praveen
    Hi David, I was just referring to the previous discussion to give better understanding to my question:) Thanks a lot for your time and patience. Praveen
    Replies:
    21
    Views:
    381
  26. Nicole Seaman

    P1.T2.405. Distributions I

    HI @theproman23 There is no sample so there is no standard error; question 405.1 is just asking about the properties of the given distribution. To contrast, let me ask a question that does invoke the standard error (which, in this case, is the standard deviation of a sample mean not a population). Here is the alternate question just for contrast: Assume a population with mean earnings of $2.5...
    HI @theproman23 There is no sample so there is no standard error; question 405.1 is just asking about the properties of the given distribution. To contrast, let me ask a question that does invoke the standard error (which, in this case, is the standard deviation of a sample mean not a population). Here is the alternate question just for contrast: Assume a population with mean earnings of $2.5...
    HI @theproman23 There is no sample so there is no standard error; question 405.1 is just asking about the properties of the given distribution. To contrast, let me ask a question that does invoke the standard error (which, in this case, is the standard deviation of a sample mean not a...
    HI @theproman23 There is no sample so there is no standard error; question 405.1 is just asking about the properties of the given distribution. To contrast, let me ask a question that does invoke...
    Replies:
    14
    Views:
    379
  27. David Harper CFA FRM

    L1.T2.109 EWMA covariance

    Hi @FM22 From Hull 23.7:
    Hi @FM22 From Hull 23.7:
    Hi @FM22 From Hull 23.7:
    Hi @FM22 From Hull 23.7:
    Replies:
    9
    Views:
    372
  28. jakub

    Q 206: Variance of sample average

    Yes, only because we are concerned with the (sample) distribution of an AVERAGE return over 5 years. This variable itself (an average of 5 variables) has a standard deviation called the sample average (If you have Hull, it is really the same as his Example 14.3 in Chapter 5), thanks
    Yes, only because we are concerned with the (sample) distribution of an AVERAGE return over 5 years. This variable itself (an average of 5 variables) has a standard deviation called the sample average (If you have Hull, it is really the same as his Example 14.3 in Chapter 5), thanks
    Yes, only because we are concerned with the (sample) distribution of an AVERAGE return over 5 years. This variable itself (an average of 5 variables) has a standard deviation called the sample average (If you have Hull, it is really the same as his Example 14.3 in Chapter 5), thanks
    Yes, only because we are concerned with the (sample) distribution of an AVERAGE return over 5 years. This variable itself (an average of 5 variables) has a standard deviation called the sample...
    Replies:
    3
    Views:
    361
  29. Suzanne Evans

    P1.T2.204. Joint, marginal, and conditional probability functions (Stock & Watson)

    Hi Melody (@superpocoyo ) Here is the spreadsheet @ Please note that, in my response to mastvikas above, I had a typo which I've now corrected. It should read: (10 - 29.38)^2*(0.05/.32) = 58.65 105.859 is the conditional variance which determines the answer of 10.3 (the conditional standard deviation). I think the key here is to realize that, after we grok the conditionality, we are...
    Hi Melody (@superpocoyo ) Here is the spreadsheet @ Please note that, in my response to mastvikas above, I had a typo which I've now corrected. It should read: (10 - 29.38)^2*(0.05/.32) = 58.65 105.859 is the conditional variance which determines the answer of 10.3 (the conditional standard deviation). I think the key here is to realize that, after we grok the conditionality, we are...
    Hi Melody (@superpocoyo ) Here is the spreadsheet @ Please note that, in my response to mastvikas above, I had a typo which I've now corrected. It should read: (10 - 29.38)^2*(0.05/.32) = 58.65 105.859 is the conditional variance which determines the answer of 10.3 (the conditional standard...
    Hi Melody (@superpocoyo ) Here is the spreadsheet @ Please note that, in my response to mastvikas above, I had a typo which I've now corrected. It should read: (10 - 29.38)^2*(0.05/.32) =...
    Replies:
    10
    Views:
    352
  30. David Harper CFA FRM

    L1.T2.72 Student's t distribution

    Hi SheldonZ, the df does not enter the calculation of the test statistic. Its calculated as: t = (x -mu) * sqrt(n)/ s where s is the sample standard deviation. The df comes into play at determining the critical value from the t - distribution, which you than use to compare it to the t - statistics from above, but that is not part of this exercise. I hope that helped. Addendum: Sometimes the...
    Hi SheldonZ, the df does not enter the calculation of the test statistic. Its calculated as: t = (x -mu) * sqrt(n)/ s where s is the sample standard deviation. The df comes into play at determining the critical value from the t - distribution, which you than use to compare it to the t - statistics from above, but that is not part of this exercise. I hope that helped. Addendum: Sometimes the...
    Hi SheldonZ, the df does not enter the calculation of the test statistic. Its calculated as: t = (x -mu) * sqrt(n)/ s where s is the sample standard deviation. The df comes into play at determining the critical value from the t - distribution, which you than use to compare it to the t -...
    Hi SheldonZ, the df does not enter the calculation of the test statistic. Its calculated as: t = (x -mu) * sqrt(n)/ s where s is the sample standard deviation. The df comes into play at...
    Replies:
    34
    Views:
    343

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