# How does standard deviation relate to probability?

This is a bank loan problem. There are 1000 loans in the pool and each loan is for 180,000 dollars. The probability of default is 0.02. The amount of money the bank loses on a loan default is 120,000. The amount of money the bank makes on the loan in 1 year is an interest rate i * 180,000.

The distribution I'm starting with has a mean of 100 and a standard deviation of 17165.73. The goal of the problem is the raise the interest rate until the probability that the bank loses money is 0.001. With the current distribution, the probability of losing money is ~50%.

Since the loan pool only has 2 types of loans (defaulting and non defaulting loans), we can use the standard deviation formula as follows:

$| a - b | * \sqrt{p (1 - p)}$ = standard deviation where

a = profitable loan = interest rate * 180,000, b = defaulting loan = -120000, p = 0.02, 1 - p = .98

So I end up with something like this $| 180,000 * i - -120,000 | * \sqrt{0.02 * (1 - 0.02)}$

However, I'm not sure where to go from here because I'm not sure how to relate the standard deviation to the desired probability outcome. If I knew what standard deviation I needed, I could set the above equation equal to that SD and solve for i. Any help in getting me going in the right direction would be appreciated. It's totally acceptable for me to use computer software like R to assist in the calculations.

## 1 Answer

I might be missing something, but I don't really understand where you are getting the mean and standard deviation... they seem to be red herrings actually.

In a single loan, as long as $i > 0$ the probability for the bank of losing money is always $0.2$.

With two loans, you have the following possible mutually-exclusive events:

• both are good: no money is lost, with probability $(1-p)^2$
• first is good, second is bad: money is gained or lost depending on $i$, with probability $p(1-p)$
• first is bad, second is good: same as before with same probability
• both are bad: money is lost, with probability $p^2$

The only way for $i$ to be relevant is on the two middle terms, that add up to a probability of $2 p (1-p)$. In particular, in these two cases the outcome would be:

$$G_{1} = 180000 i - 120000$$

so if we solve for $G_1 = 0$ we get $i = 2/3$. Hence:

• if $i \geq 2/3$, the middle term is not a loss (it's either 0, or a gain), and the probability of a loss is $p^2$ (i.e. both loans default)
• otherwise, if $i < 2/3$, the middle term is a loss, and the probability of a loss is $p^2 + 2p(1-p)$

The average gain would be given by:

$$G = G_0 (1-p)^2 + 2 G_1 p (1-p) + G_2 p^2$$

where $G_i$ is the outcome in case of $i$ defaults (0, 1 or 2 in this case). As we saw, the intermediate term $G_1$ can contribute as a loss term or not depending on $i$.

Generalizing for the case of N loans:

$$G = \sum_{ 0 \leq d \leq N}{\binom{N}{d} G_d p^d (1-p)^{N-d}}$$

As we saw, the term with $d = 0$ is always a gain as long as $i > 0$, while the term with $d = N$ is always a loss, independently of $i$. The intermediate term can be a gain or a loss depending on the sign of $G_d$:

$$G_d = (N - d) 180000 i - 120000 d$$

For each possible value $d^*$ from $0$ to $N$, you can calculate a value $i_{d^*}$ that makes $G_{d^*}$ equal to zero:

$$i_{d^*} = \frac{120000 d^*}{180000 (N - d^*)}$$

For this choice of $i$, all terms $G_d$ with $d > d^*$ will be contributing to a loss, and all other terms will not (either gain, or zero). Actually, this happens for any choice of $i$ that satisfies:

$$i_{d^*} \leq i \lt i_{d^* + 1}$$

In other terms, $d^*$ represents the maximum number of defaults you can withstand without suffering a loss. Any number of defaults higher than this will yield a loss.

Each term is associated to a probability that is given by:

$$p_d = \binom{N}{d} p^d (1-p)^{N-d}$$

so for $i$ in the range above the probability of having a loss is the sum of all $p_d$ for $d > d^*$:

$$P_{loss}(d^*) = \sum_{d^* < d \leq N}{\binom{N}{d} p^d (1-p)^{N-d} }$$

Now, there might be some clever formula for doing this calculation, but I'm not really into this field and I'm too lazy. From an algorithmic point of view, I'd do as follows:

• calculate $D$ such that $P_{loss}(D) \leq 0.001 \lt P_{loss}(D - 1)$
• calculate $i_{D}$ and $i_{D + 1}$
• choose any value $i$ such that $i_{D} \leq i \lt i_{D + 1}$, depending on the greed of the bank.

I hope I didn't put too many errors...