# P1.T2. Quantitative Analysis

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

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1. ### 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:
1,583
2. ### 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. ### 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
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51
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957
4. ### 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...
Pam Gordon ... 2 3
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43
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920
5. ### 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...
Pam Gordon ... 2 3
Replies:
41
Views:
815
6. ### 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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42
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807
7. ### 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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791
8. ### 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...
Replies:
43
Views:
782
9. ### 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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40
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705
10. ### 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...
Fran ... 2
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21
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652
11. ### P1.T2.303 Mean and variance of continuous probability density functions (pdf)

Concept cleared, thanks
Concept cleared, thanks
Concept cleared, thanks
Concept cleared, thanks
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38
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643
12. ### 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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34
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605
13. ### 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
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26
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588
14. ### 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...
Fran ... 2
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583
15. ### 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. I hope I get a smile looking at my FRM result !
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. I hope I get a smile looking at my FRM result !
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. I hope I get a smile looking at my FRM result !
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. I hope I get a smile looking at my FRM result !
LL
Replies:
2
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564
16. ### 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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20
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557
17. ### 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. ### 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]...
Fran ... 2
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495
19. ### 63.1

Thanks David!
Thanks David!
Thanks David!
Thanks David!
LL
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12
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477