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)

    Hi @bake5472 Yes, exactly! I still don't see why Tosuhn's shortcut works .... We are given an upper domain bound of e^5 and he directly retrieves the answer with e^(95%*5), by what logic? You are correct in your "proof." As I wrote above at , in order to qualify as a probability distribution, by definition the pdf must integrate to zero such that: ...
    Hi @bake5472 Yes, exactly! I still don't see why Tosuhn's shortcut works .... We are given an upper domain bound of e^5 and he directly retrieves the answer with e^(95%*5), by what logic? You are correct in your "proof." As I wrote above at , in order to qualify as a probability distribution, by definition the pdf must integrate to zero such that: ...
    Hi @bake5472 Yes, exactly! I still don't see why Tosuhn's shortcut works .... We are given an upper domain bound of e^5 and he directly retrieves the answer with e^(95%*5), by what logic? You are correct in your "proof." As I wrote above at , in order to qualify as a probability distribution, by...
    Hi @bake5472 Yes, exactly! I still don't see why Tosuhn's shortcut works .... We are given an upper domain bound of e^5 and he directly retrieves the answer with e^(95%*5), by what logic? You are...
    Replies:
    53
    Views:
    1,545
  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:
    957
  4. Pam Gordon

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

    Hi @David Harper CFA FRM, thanks for taking the time to reply :). There's one tweak I would like to apply to this problem - let's say instead of a 12 step binomial model, we have a 8 step binomial and the rest of the info staying the same. We would now have 5 ups and 3 downs to reach $121 right? And the binomial probability will be given by: Binomial Probability [X = 5 | n = 8, p =60%] = 27.89%?
    Hi @David Harper CFA FRM, thanks for taking the time to reply :). There's one tweak I would like to apply to this problem - let's say instead of a 12 step binomial model, we have a 8 step binomial and the rest of the info staying the same. We would now have 5 ups and 3 downs to reach $121 right? And the binomial probability will be given by: Binomial Probability [X = 5 | n = 8, p =60%] = 27.89%?
    Hi @David Harper CFA FRM, thanks for taking the time to reply :). There's one tweak I would like to apply to this problem - let's say instead of a 12 step binomial model, we have a 8 step binomial and the rest of the info staying the same. We would now have 5 ups and 3 downs to reach $121...
    Hi @David Harper CFA FRM, thanks for taking the time to reply :). There's one tweak I would like to apply to this problem - let's say instead of a 12 step binomial model, we have a 8 step binomial...
    Replies:
    43
    Views:
    906
  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:
    803
  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!
    Replies:
    42
    Views:
    801
  7. 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
  8. Nicole Manley

    P1.T2.312. Mixture distributions

    Hi @ami44 oh, right! I think I understand how they are the same. Sorry, I did not notice the greater sign (">") in "s^2 >= w1 * s1^2 + w2 * s2^2." That is very informative, I did not know this!
    Hi @ami44 oh, right! I think I understand how they are the same. Sorry, I did not notice the greater sign (">") in "s^2 >= w1 * s1^2 + w2 * s2^2." That is very informative, I did not know this!
    Hi @ami44 oh, right! I think I understand how they are the same. Sorry, I did not notice the greater sign (">") in "s^2 >= w1 * s1^2 + w2 * s2^2." That is very informative, I did not know this!
    Hi @ami44 oh, right! I think I understand how they are the same. Sorry, I did not notice the greater sign (">") in "s^2 >= w1 * s1^2 + w2 * s2^2." That is very informative, I did not know this!
    Replies:
    40
    Views:
    750
  9. 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:
    705
  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
    Views:
    634
  11. Suzanne Evans

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

    Sweet! Keep the questions coming!
    Sweet! Keep the questions coming!
    Sweet! Keep the questions coming!
    Sweet! Keep the questions coming!
    Replies:
    37
    Views:
    616
  12. 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
    Views:
    602
  13. 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:
    564
  14. Nicole Manley

    P1.T2.504. Copulas (Hull)

    Hello @hellohi These learning objectives are included in the 2016 GARP curriculum under Topic 2, Reading 14, Hull Chapter 11. Nicole
    Hello @hellohi These learning objectives are included in the 2016 GARP curriculum under Topic 2, Reading 14, Hull Chapter 11. Nicole
    Hello @hellohi These learning objectives are included in the 2016 GARP curriculum under Topic 2, Reading 14, Hull Chapter 11. Nicole
    Hello @hellohi These learning objectives are included in the 2016 GARP curriculum under Topic 2, Reading 14, Hull Chapter 11. Nicole
    Replies:
    26
    Views:
    561
  15. 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:
    547
  16. Fran

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

    Hi @hellohi Yes, that's correct. In credit risk analysis, "default" is typically classified with a 1, while "no default" is classified as a zero; my answer happens to retain this conventional usage such that P(default) = P(X = 1) = 5% and P(no default; aka, survive) = P(X = 0 ) = 95%. However, as you might expect, switching these do not really alter the outcome except to produce a negative...
    Hi @hellohi Yes, that's correct. In credit risk analysis, "default" is typically classified with a 1, while "no default" is classified as a zero; my answer happens to retain this conventional usage such that P(default) = P(X = 1) = 5% and P(no default; aka, survive) = P(X = 0 ) = 95%. However, as you might expect, switching these do not really alter the outcome except to produce a negative...
    Hi @hellohi Yes, that's correct. In credit risk analysis, "default" is typically classified with a 1, while "no default" is classified as a zero; my answer happens to retain this conventional usage such that P(default) = P(X = 1) = 5% and P(no default; aka, survive) = P(X = 0 ) = 95%. However,...
    Hi @hellohi Yes, that's correct. In credit risk analysis, "default" is typically classified with a 1, while "no default" is classified as a zero; my answer happens to retain this conventional...
    Replies:
    19
    Views:
    540
  17. 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:
    520
  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
    Views:
    486
  19. LL

    63.1

    Thanks David!
    Thanks David!
    Thanks David!
    Thanks David!
    Replies:
    12
    Views:
    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
    Views:
    416

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