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)

    @PortoMarco79 I received a response to my question .... Bill May replied:
    @PortoMarco79 I received a response to my question .... Bill May replied:
    @PortoMarco79 I received a response to my question .... Bill May replied:
    @PortoMarco79 I received a response to my question .... Bill May replied:
    Replies:
    42
    Views:
    1,343
  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:
    1,027
  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:
    939
  4. 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
  5. Pam Gordon

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

    Thanks!, David!
    Thanks!, David!
    Thanks!, David!
    Thanks!, David!
    Replies:
    35
    Views:
    782
  6. Suzanne Evans

    P1.T2.209 T-statistic and confidence interval

    Hi @bpdulog yes that is an accurate criticism: it does need to be a random sample. You are right. At the same time, it's a safe assumption to make if "random" happens to be omitted, as in GARP's 2016 Practice P1 Question 64: "For a sample of the past 30 monthly stock returns for McCreary, Inc., the mean return is 4% and the sample standard deviation is 20%. Since the population variance is...
    Hi @bpdulog yes that is an accurate criticism: it does need to be a random sample. You are right. At the same time, it's a safe assumption to make if "random" happens to be omitted, as in GARP's 2016 Practice P1 Question 64: "For a sample of the past 30 monthly stock returns for McCreary, Inc., the mean return is 4% and the sample standard deviation is 20%. Since the population variance is...
    Hi @bpdulog yes that is an accurate criticism: it does need to be a random sample. You are right. At the same time, it's a safe assumption to make if "random" happens to be omitted, as in GARP's 2016 Practice P1 Question 64: "For a sample of the past 30 monthly stock returns for McCreary, Inc.,...
    Hi @bpdulog yes that is an accurate criticism: it does need to be a random sample. You are right. At the same time, it's a safe assumption to make if "random" happens to be omitted, as in GARP's...
    Replies:
    38
    Views:
    708
  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:
    707
  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
    Views:
    701
  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:
    35
    Views:
    619
  10. Suzanne Evans

    P1.T2.212. Difference between two means

    Thank you so much!!!
    Thank you so much!!!
    Thank you so much!!!
    Thank you so much!!!
    Replies:
    29
    Views:
    566
  11. 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
    Views:
    560
  12. 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:
    24
    Views:
    527
  13. 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:
    20
    Views:
    524
  14. 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:
    505
  15. Fran

    P1.T2.301. Miller's probability matrix

    David, you're the best! This definitely helps. And thanks for detailed response.
    David, you're the best! This definitely helps. And thanks for detailed response.
    David, you're the best! This definitely helps. And thanks for detailed response.
    David, you're the best! This definitely helps. And thanks for detailed response.
    Replies:
    13
    Views:
    502
  16. 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,...
    Replies:
    17
    Views:
    490
  17. 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:
    22
    Views:
    490
  18. LL

    63.1

    Thanks David!
    Thanks David!
    Thanks David!
    Thanks David!
    Replies:
    12
    Views:
    477
  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]...
    Replies:
    22
    Views:
    471
  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
    Views:
    389

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