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)

    yes you are right my fault was here p(n) = n! /m!*(m-n)! * p^m * (1-p)^(n-m) thanks dear, now we are both correct thanks for help ;)
    yes you are right my fault was here p(n) = n! /m!*(m-n)! * p^m * (1-p)^(n-m) thanks dear, now we are both correct thanks for help ;)
    yes you are right my fault was here p(n) = n! /m!*(m-n)! * p^m * (1-p)^(n-m) thanks dear, now we are both correct thanks for help ;)
    yes you are right my fault was here p(n) = n! /m!*(m-n)! * p^m * (1-p)^(n-m) thanks dear, now we are both correct thanks for help ;)
    Replies:
    51
    Views:
    1,458
  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
    Views:
    1,028
  3. David Harper CFA FRM

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

    Hi David, Disregard, got it now Regards Alex
    Hi David, Disregard, got it now Regards Alex
    Hi David, Disregard, got it now Regards Alex
    Hi David, Disregard, got it now Regards Alex
    Replies:
    51
    Views:
    948
  4. Pam Gordon

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

    Hi @David Harper CFA FRM, thank you for that clarification, it was something that I was came up with! Every time I read your post I understand that I need to know so much things. I am glad that I got a opportunity to learn from you:). I appreciate your help and clarification. Thank you :)
    Hi @David Harper CFA FRM, thank you for that clarification, it was something that I was came up with! Every time I read your post I understand that I need to know so much things. I am glad that I got a opportunity to learn from you:). I appreciate your help and clarification. Thank you :)
    Hi @David Harper CFA FRM, thank you for that clarification, it was something that I was came up with! Every time I read your post I understand that I need to know so much things. I am glad that I got a opportunity to learn from you:). I appreciate your help and clarification. Thank you :)
    Hi @David Harper CFA FRM, thank you for that clarification, it was something that I was came up with! Every time I read your post I understand that I need to know so much things. I am glad that I...
    Replies:
    42
    Views:
    864
  5. 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
    Views:
    791
  6. Suzanne Evans

    P1.T2.209 T-statistic and confidence interval

    Hi @FRMeugene no worries, I'm proud of this question but it's tricky. Let's solve it another way, by starting with your correct statement that "isn't the variance of a binomial just [n*p*(1-p)]?" Yes, true (). So if p = pd = 15% and n = 60, then variance = 15%*(1-15%)*60 = 7.65. Please note this is a variance of 7.65 defaults^2 so the standard deviation of this distribution of 60 companies is...
    Hi @FRMeugene no worries, I'm proud of this question but it's tricky. Let's solve it another way, by starting with your correct statement that "isn't the variance of a binomial just [n*p*(1-p)]?" Yes, true (). So if p = pd = 15% and n = 60, then variance = 15%*(1-15%)*60 = 7.65. Please note this is a variance of 7.65 defaults^2 so the standard deviation of this distribution of 60 companies is...
    Hi @FRMeugene no worries, I'm proud of this question but it's tricky. Let's solve it another way, by starting with your correct statement that "isn't the variance of a binomial just [n*p*(1-p)]?" Yes, true (). So if p = pd = 15% and n = 60, then variance = 15%*(1-15%)*60 = 7.65. Please note this...
    Hi @FRMeugene no worries, I'm proud of this question but it's tricky. Let's solve it another way, by starting with your correct statement that "isn't the variance of a binomial just [n*p*(1-p)]?"...
    Replies:
    40
    Views:
    761
  7. Pam Gordon

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

    Hi Olga (@Ljula777 ) I'll assume a variant of 310.1 where the bond contains 100 obligors with average default rate of 4.0%, such that the mean of a Poisson, lambda (λ), is given by 4.0. Then the Prob[X = 6] = λ^x*e^(-λ)/x! = 4^6*e^(-4)/6! and, with my TI BAII+ calculator, my steps are: Steps: ...................... Display shows: 4[y^x]6[×] ..........................
    Hi Olga (@Ljula777 ) I'll assume a variant of 310.1 where the bond contains 100 obligors with average default rate of 4.0%, such that the mean of a Poisson, lambda (λ), is given by 4.0. Then the Prob[X = 6] = λ^x*e^(-λ)/x! = 4^6*e^(-4)/6! and, with my TI BAII+ calculator, my steps are: Steps: ...................... Display shows: 4[y^x]6[×] ..........................
    Hi Olga (@Ljula777 ) I'll assume a variant of 310.1 where the bond contains 100 obligors with average default rate of 4.0%, such that the mean of a Poisson, lambda (λ), is given by 4.0. Then the Prob[X = 6] = λ^x*e^(-λ)/x! = 4^6*e^(-4)/6! and, with my TI BAII+ calculator, my steps are: Steps: ...
    Hi Olga (@Ljula777 ) I'll assume a variant of 310.1 where the bond contains 100 obligors with average default rate of 4.0%, such that the mean of a Poisson, lambda (λ), is given by 4.0. Then the...
    Replies:
    39
    Views:
    738
  8. 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
    Replies:
    40
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    703
  9. Nicole Manley

    P1.T2.312. Mixture distributions

    Thanks, that is helpful. When standardizing, is the population mean always the one being subtracted?
    Thanks, that is helpful. When standardizing, is the population mean always the one being subtracted?
    Thanks, that is helpful. When standardizing, is the population mean always the one being subtracted?
    Thanks, that is helpful. When standardizing, is the population mean always the one being subtracted?
    Replies:
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    656
  10. Fran

    P1.T2.301. Miller's probability matrix

    HI @ami44 Those are really thoughtful points, thank you. Yes, I do agree with your first point: default is a random variable characterized by a Bernoulli distribution (i.e., two discrete outcomes). The default probability (PD; aka, EDF) is really the mean (expected value) of the Bernoulli such that PD = E(X) = Prob(default) or P(X = 1) Similarly, my LGD is imprecise (at best). As you say,...
    HI @ami44 Those are really thoughtful points, thank you. Yes, I do agree with your first point: default is a random variable characterized by a Bernoulli distribution (i.e., two discrete outcomes). The default probability (PD; aka, EDF) is really the mean (expected value) of the Bernoulli such that PD = E(X) = Prob(default) or P(X = 1) Similarly, my LGD is imprecise (at best). As you say,...
    HI @ami44 Those are really thoughtful points, thank you. Yes, I do agree with your first point: default is a random variable characterized by a Bernoulli distribution (i.e., two discrete outcomes). The default probability (PD; aka, EDF) is really the mean (expected value) of the Bernoulli such...
    HI @ami44 Those are really thoughtful points, thank you. Yes, I do agree with your first point: default is a random variable characterized by a Bernoulli distribution (i.e., two discrete...
    Replies:
    21
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    612
  11. 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!
    Replies:
    34
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    599
  12. 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:
    Replies:
    2
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    561
  13. Suzanne Evans

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

    Hi @bpdulog Because several candidates were asking, last month we asked Bill May, SVP at GARP, the following question. My question and his response: Response from Bill May (March 8, 2016):
    Hi @bpdulog Because several candidates were asking, last month we asked Bill May, SVP at GARP, the following question. My question and his response: Response from Bill May (March 8, 2016):
    Hi @bpdulog Because several candidates were asking, last month we asked Bill May, SVP at GARP, the following question. My question and his response: Response from Bill May (March 8, 2016):
    Hi @bpdulog Because several candidates were asking, last month we asked Bill May, SVP at GARP, the following question. My question and his response: Response from Bill May (March 8, 2016):
    Replies:
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    541
  14. 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)
    Replies:
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    541
  15. 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...
    Replies:
    27
    Views:
    512
  16. Nicole Manley

    P1.T2.504. Copulas (Hull)

    Looks like this was covered here:
    Looks like this was covered here:
    Looks like this was covered here:
    Looks like this was covered here:
    Replies:
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    512
  17. Fran

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

    Hi @puneet_ For this bond (a Bernoulli) we have Pr[X=0] = 95% and P[X=1] = 1- 95% = 5%. The mean, µ = (95%*0)+(5%*1) = 0.050. Skew is the third central moment divided by (ie, standardized by) σ^3, just as kurtosis is the fourth central moment divided by σ^4. With respect to skew, the third central moment = (1.0 outcome - 0.05 mean µ )^3*5% probability of 1.0 + (0 outcome - 0.05 mean µ )^3*95%...
    Hi @puneet_ For this bond (a Bernoulli) we have Pr[X=0] = 95% and P[X=1] = 1- 95% = 5%. The mean, µ = (95%*0)+(5%*1) = 0.050. Skew is the third central moment divided by (ie, standardized by) σ^3, just as kurtosis is the fourth central moment divided by σ^4. With respect to skew, the third central moment = (1.0 outcome - 0.05 mean µ )^3*5% probability of 1.0 + (0 outcome - 0.05 mean µ )^3*95%...
    Hi @puneet_ For this bond (a Bernoulli) we have Pr[X=0] = 95% and P[X=1] = 1- 95% = 5%. The mean, µ = (95%*0)+(5%*1) = 0.050. Skew is the third central moment divided by (ie, standardized by) σ^3, just as kurtosis is the fourth central moment divided by σ^4. With respect to skew, the third...
    Hi @puneet_ For this bond (a Bernoulli) we have Pr[X=0] = 95% and P[X=1] = 1- 95% = 5%. The mean, µ = (95%*0)+(5%*1) = 0.050. Skew is the third central moment divided by (ie, standardized by) σ^3,...
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    17
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    503
  18. 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]...
    Replies:
    22
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    478
  19. LL

    63.1

    Thanks David!
    Thanks David!
    Thanks David!
    Thanks David!
    Replies:
    12
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    477
  20. Fran

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

    Hi @bpdulog Please see
    Hi @bpdulog Please see
    Hi @bpdulog Please see
    Hi @bpdulog Please see
    Replies:
    8
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    408

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