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P1.T2.20.1. Conditionally independent events

Nicole Seaman

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Learning objectives: Describe an event and an event space. Describe independent events and mutually exclusive events. Explain the difference between independent events and conditionally independent events.

Questions:

20.1.1. A specialized credit portfolio contains only three loans but they are very risky, as each has a single-period default probability of 10.0%. They are independent (therefore, we have the i.i.d. condition). You know enough probability to determine (for example) that, at the end of a single period, the probability that all three loans default is 0.1% and the probability that all three loans survive is 72.9%. However, at the end of the period, the portfolio manager gives you a piece of additional information when she tells you that "AT LEAST two of the bonds have defaulted." What is the (conditional) probability that the other (third) bond also defaulted?

a. 0.09%
b. 0.10%
c. 3.57%
d. 10.0%


20.1.2. Yesterday a web page hosted by Acme received tens of thousands of page views but some were views by malicious bots. Acme utilizes two software applications to detect these malicious "bot-views." It uploads the same data file from yesterday to both applications. The first application detects 200 bot-views and the second application detects 300 bot-views. Among these, only 40 bot-views were detected by both applications. All bot-views are equally likely to be located, but clearly both applications only identify a minority of the bot-views (otherwise there would be a much higher number of identified bot-views common to both applications). Further, the identification of a bot-view by one application is independent of its identification by the other application. How many malicious bot-views did the web page experience on this day?

a. 300
b. 460
c. 540
d. 1,500


20.1.3. Albert and Betty share an office where each month they attempt to predict the best-performing industry within their respective sectors. Albert's sector is Financials and Betty's sector is Information Technology. Each contains several industries. Without any help, the probability that Albert predicts the best-performing industry (within Financials) is 12.0%, and the probability that Betty predicts the best-performing industry (within I.T.) is 15.0%. Put another way, their unconditional success probabilities are, respectively, P(A) = 12.0% and P(B) = 15.0%. Without any help, the probability that they both simultaneously predict their best industry is 1.80%; that is, the joint Pr(A ∩ B) = 1.80%. Their firm also subscribes to software with artificial intelligence and the software boosts their predictive abilities. In fact, when using the software to help them, their respective success probabilities double. Specifically, P(A | S) = 24.0% and P(B | S) = 30.0%; for example, the probability that Betty picks the best-performing industry conditional on her utilization of the software jumps to 30.0%. When they both use the software, their joint probability of success is 15.0%. In regard to the observed dependencies, which of the following statements is accurate?

a. Independent and conditionally independent
b. Independent but conditionally dependent
c. Dependent but conditionally independent
d. Dependent and conditionally dependent

Answers here:
 
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WarrenSV

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#2
Hi Guys,

Really struggling with the method to get to this answer. In particular the first part of the answer (unconditional probability) - highlighted in red below. Any further explanation available?

Q: A specialized credit portfolio contains only three loans but they are very risky, as each has a single-period default probability of 10.0%. They are independent (therefore, we have the i.i.d. condition). You know enough probability to determine (for example) that, at the end of a single period, the probability that all three loans default is 0.1% and the probability that all three loans survive is 72.9%. However, at the end of the period, the portfolio manager gives you a piece of additional information when she tells you that "AT LEAST two of the bonds have defaulted." What is the (conditional) probability that the other (third) bond also defaulted?
a) 0.09%
b) 0.10%
c) 3.57%
d) 10.0%

A: C. True: 3.57% The unconditional probability that TWO or MORE loans default equals 3*(10%^2*90%) + 10%^3 = 2.80% such that the conditional probability, Pr (3 default | two or more default) = 0.10% / 2.80% = 3.5714%.
 

David Harper CFA FRM

David Harper CFA FRM
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#3
Hi @WarrenSV In the source at https://www.bionicturtle.com/forum/threads/p1-t2-20-1-conditionally-independent-events.23249/ I explained ...
Hi @carroleo Because there is only one permutation of three defaults but there are three permutations of two defaults. In question 20.1.1, where we have a portfolio of three loans and each an either default or survive, there are 2^3 = 8 outcomes, as follows (where S = survive and D = default):

{S, S, S}, {S, S, D}, {S, D, S}, {D, S, S}, {S, D, D}, {D, S, D}, {D, D, S}, {D, D, D}

The probability of each of these outcomes is given by multiplication; e.g., Pr({D, S, S}) = 10%*90%*90% = 8.1%; Pr({DSD) = 10%*90%*10% = 0.90%. We can confirm that adding up the probability of each outcome is 100.0% (aka, mutually exclusive cumulatively exhaustive). These outcomes map to the four events. There are four events:
  • zero defaults: {S, S, S} with Prob = 90%^3 = 72.90%
  • one default: {S, S, D}, {S, D, S}, {D, S, S} with Prob = 8.1% + 8.1% + 8.1% = 24.30%
  • two defaults: {S, D, D}, {D, S, D}, {D, D, S} with Prob = 0.90% + 0.90% + 0.90% = 2.70%
  • three defaults: {D, D, D} with Prob = 10%^3 = 0.10%
The sum of these must also be 100.0%. So, the 10.0%^2*90*3 is because there are three ways (outcomes) to get one default: either the first or second or third bond defaults. But there is only one way for all three bonds to default. I hope that's helpful!
... so this is just a binomial distribution (btw, this is easily the most common credit distribution in the FRM just because it's so tractable). Because we have three bonds (n = 3) with PD = p = 10%, the binomial pmf (where X is the number of defaults) is this:

The unconditional probability of two or more defaults is the probability of an outcome of 2 or 3; i.e., 2.7% + 0.1% = 2.8%. When I wrote "3*(10%^2*90%) + 10%^3 = 2.80%" I was using the binomial pmf if we want to take the long way:
  • Pr(X = 2 defaults) = C(n,k)*p^k*(1-p)^(n-k) = C(3,2) * 10%^2 * 90%^1 = 3 * 10%^2 * 90% = 2.70%, and
  • Pr(X = 3 defaults) = C(n,k)*p^k*(1-p)^(n-k) = C(3,3) * 10%^3 * 90%^0 = 1 * 10%^3 * 1.0 = 10%^3 = 0.1%
You just don't need the whole formula, or really more to the point, it's best to understand what the binomial is doing so we don't blindly misapply it. The probability of two defaults and one survival is 10%^2*90% and there are three combinations, {S, D, D}, {D, S, D}, {D, D, S}, per my itemization above. In the binomial, C(n,K) is the binomial coefficient (https://en.wikipedia.org/wiki/Binomial_coefficient); in this case, C(3,2) = COMBIN(3,2) because there are 3 ways to choose 2 items from a set of 3.

By the way, while I'm here, I'll add two academic points. First, if we want to think of bond survival (non-default) as binomial success, that's fine. Then p = 90% and the pmf is:


My first version is L(+)/P(-) or L/P format and how we typically render risk distributions: losses to the right, but that can be confusing because it switches the normal mathematical scale we are used to. We're used to negatives on the left. So my second version is P(+)/L(-) or P/L because, on that scale X = 0 is three defaults and therefore losses of 3 * Face * LGD; this is familar to us, but maybe less common in risk for a few reasons. Neither is wrong.

My second academic point is: notice that the answer to the question, "What is the probability that the third bond defaulted conditional on being told by the PM that two or more bonds defaulted?" is neither, as we might expect, the unconditional Bernouilli (n = 1 trial) 10.0% probability; nor is it the (unconditional) probability that all three bonds default.

I think it's natural to guess that the answer to 20.1.1. is 10% because, if we have three bonds, the unconditional probability that the third bond defaults is 10%. This can be seen if we itemize, from among the total of eight outcomes, the four that include the third bond defaulting:

{S, S, S}
{S, S, D} --> 8.10%
{S, D, S}
{D, S, S} -->
{S, D, D} --> 0.90%
{D, S, D} --> 0.90%
{D, D, S}
{D, D, D} -->0.10%
The sum = 8.10% + 0.90% + 0.90% + 0.10% = 10.0%. But we don't need to go to this trouble (!!) because already know the probability of a single bond's default is 10.0%. The point I want to make is that intuitively we know the 3rd bond's outcome is independent of the other two. But this is answer to a different question, it answers the alternative question, what is Pr(3rd bond defaults | first two default)? Here is the point: if these two events are independent, then Pr(3rd bond defaults | first two default) = Pr(3rd bond defaults), and indeed, Pr(3rd bond defaults | first two default) = 0.10% / (0.10% + 0.90%) = 0.10% / 1.0% = 10.0%.

Nor is the answer Pr(X = 3) = 10%^3 = 0.1%. The actual conditional question is, what is Pr ( 3 defaults | 2 or more defaults) or, to use easier notation, let's say Pr(3D | 2D+). If these two events, 3D and 2+, are independent then Pr(3D)*Pr(2D+) = P(3D ∩ 2D+) and Pr(3D) = P(3D ∩ 2D+)/Pr(2D+) = Pr(3D | 2D+). That's the point, if these events are independent, then unconditional Pr(3D) = conditional Pr(3D | 2D+). But Pr(3D) = 0.10% and, per the answer Pr(3D | 2D+) = 0.10% / 2.80% = 3.5714% because these events are not independent. The product Pr(3D)*Pr(2D+) = 0.10%*2.80% = 0.0028%, but the joint product P(3D ∩ 2D+) is actually 0.10%. The only reason I did this is to illustrate the property of independence. In order to justify an answer of 10.0%, we would have to assume independence between the two events (that constitute the question), but they are not independent!

That is a long-winded way of saying that:
  • The question asked is, what is the conditional Pr(3D | 2D+)?
  • It is not, what is the conditional Pr(3rd bond defaults | first two default)?
  • Nor is it, what is the unconditional Pr(all three bonds default)?
My inspiration for the question was a simpler series of questions:
Problem:
(a) You bump into a random person on the street who says, “I have two children. At least one of them is a boy.” What is the probability that the other child is also a boy?
(b) You bump into a random person on the street who says, “I have two children. The older one is a boy.” What is the probability that the other child is also a boy?
(c) You bump into a random person on the street who says, “I have two children, one of whom is this boy standing next to me.” What is the probability that the other child is also a boy? --

Morin, David. Probability: For the Enthusiastic Beginner (p. 102). Kindle Edition.
For me, the idea is (sincerely) not to be clever or profound, it's just hopefully a fun excuse to reinforce these concepts of conditional, unconditional, joint, and indepence. I hope that's interesting!
 
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