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I have question on probability of default as it relates to a set of loans vs individual loans. I am unable to reconcile the following two viewpoints:

Let's say I have 100 loans and I observe 20 defaults and hence, I define my PD to be 20%. Basis this, I want to know what is the probability of default for a particular loan.

I came up with the following two setups:

Setup 1

I simply run some simulations and find the number of loans that default on average. Here's some R code:

N <- 100
nDef <- 20
pd <- nDef/N

nRandom <- 10000
out <- NULL

for(i in 1:nRandom){
  r <- rbinom(N, 1, pd)
  out <- rbind(out, table(r))
}

apply(out, 2, median)

The output is 0: 80, 1: 20 which supports the idea that on average I expect a particular loan to default with a probability of 20%.

Setup 2

I could think of doing this another way: (R code below)

(pd^n * (1-pd)^(N-n)) * choose(N, n)

But doing this gives me ~10% - What am I doing wrong here?

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    $\begingroup$ You have to calculate $\frac1{N}\cdot \sum\limits_{n=1}^N n\cdot \binom{N}{n}\cdot pd^n\cdot (1-pd)^{N-n}$ If $pd=0.2$ it gives $20\%$ $\endgroup$ – callculus Jun 14 '18 at 7:06
  • $\begingroup$ Thank you for your response - can I ask you this: What is the difference if one is trying to find the probability of observing 20 bad loans out of 100 vs what is the probability of say a given loan turning out to be bad? Are they conceptually the same question? Should the response in both cases be 20%? $\endgroup$ – r2k4 Jun 14 '18 at 8:17
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trying to find the probability of observing 20 bad loans out of 100

If the probability that a given loan turning out to be bad is $\text{constant}$ 20% then you can calculate the probability to get $n$ bad loans if you select $N$ loans. This can be made by using the binomial distribution.

$$P(X=n)=\binom{N}{n}\cdot 0.2^n\cdot (1-0.2)^{N-n}$$

Thus the probability to observe 20 bad loans out of 100 is

$$P(X=20)=\binom{100}{20}\cdot 0.2^{20}\cdot (0.8)^{80}\approx 9.93\%$$

But you can calculate the probability to observe 57 bad loans out of 100 as well, for instance.

what is the probability of say a given loan turning out to be bad

This constant probability is an assumption, which has to be made to apply the binomial distribution above. The condition is that you have a large number of loans (population). In this case the ratio of number of bad loans and number of all loans at the population is $20\%$. Then the probability that you draw a bad loan at every drawing is approximately $pd=0.2$, if the population is large. If the population is not large, then you have to use the $\text{hypergeometric distribution}$.

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  • $\begingroup$ One last question if I may: Let's say we have an infinite supply of loans and if I am to draw a 100 loans, on average I expect to draw 20 bad loans and 80 good loans. But if I try to compute what are the chances that I draw 20 bad loans and 80 good loans on a particular draw, that turns out to be ~10% as you have confirmed above. Why is that not close to 20% - since that what I assumed in the first place? $\endgroup$ – r2k4 Jun 14 '18 at 11:11
  • $\begingroup$ @r2k4 The 20 percent in the population has not much to do with the 20% (20 of 100) at the sample. There are many factors which influences $P(X=n)$, like $n$, $N$ and $p$. You see that you can conclude from $p$ only what the Probability of $P(X=n)$ is. There is no direct link between $p$ and $P(X=n)$ in that way you expected. But there are some properties of the binomial distribution where p is involved like the expected value, variance, etc. Just have a look at this distribution $\endgroup$ – callculus Jun 14 '18 at 11:48

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