Question

For the next two problems, there are N 10000 car drivers. Each gets into an accident with probability 15%. In case of an accident, the losses are distributed uniformly on 0,10. Assume independence. Problem 6.13. Find the probability that the total losses exceed 8000. Problem 6.14. Find the value at risk at the confidence level 95%.

Answer 1

No. of car drivers = N=10000

Suppose, $X_i$ is a random variable such that :

$X_i =\left\{ \begin{matrix} 1 & if \quad i^{th} \quad has \quad accident \\ 0 & otherwise\end{matrix} \right.\quad \forall \quad i=1(1)N$

As per question, we have ,

Since the losses are distributed uniformly on [0,10] in case of accident, if Y_idenotes a random variable which represents the losses incurred by the $i^{th}$ car driver after an accident then ,

$Pr[Y_i=y | X_i = 1] = \frac{1}{10} \quad \forall \quad 0 \leq y \leq 10 , i=1(1)N$

$Pr[Y_i=y | X_i = 0] = 0 \quad \forall \quad 1 \leq y \leq 10 , i=1(1)N$

Now,

For $1 \leq y\leq 10$

$= 0.15*\frac{1}{10}=0.015$

Thus we have,

$E(Y_i) = 0+(1+2+3+\dots +10)*0.015= \frac{10*11}{2}*0.015=0.825$

$E(Y_i^2) = 0+(1^2+2^2+3^2+\dots +10^2)*0.015= 385*0.015=5.775$

$V(Y_i)= E(Y_i^2) - \left( E(Y_i)\right)^2=5.094375$

Since we are to assume that $Y_i$ 's are independent:

$E(\sum_{i=1}^{N}Y_i)=8250 \quad \& \quad V\left( \sum_{i=1}^{N}Y_i \right )=\sum_{i=1}^{N}V(Y_i)=509000$

a) Suppose $T= \sum_{i=1}^{N}Y_i$ is the total losses, then

Probability that the total losses exceed 8000 =

$=Pr \left[ \frac{T-E(T)}{\sqrt{V(T)}} > \frac{8000-E(T)}{\sqrt{V(T)}}\right]$

$=Pr \left[ Z > -1.11 \right ]$

(where Z follows a N(0,1) distribution)

$=Pr \left[ Z \leq 1.11 \right ]$

$=0.8665$

b) Suppose we have,  $\tau$ and which follows N(0,1), such that,

$Pr \left[ |Z| \leq \tau \right ] = 0.95 \\ \implies Pr\left[ -\tau \leq Z \leq \tau \right ] = 0.95 \\ \implies 2*Pr\left[ 0 \leq Z \leq \tau \right ] =0.95 \\ \implies Pr[Z \leq \tau] -Pr[Z \leq 0]=0.475 \\ \implies Pr[Z \leq \tau]=0.975$

From the Standard Normal Table, we find that for $\tau =1.96$ , we have $Pr[|Z| \leq \tau] =0.95$ .

Thus, [-1.96 , 1.96 ] is the 95% confidence interval.

We know that,

$Z= \frac{T -E(T)}{\sqrt{V(T)}} \\ \implies T = E(T) + Z*\sqrt{V(T)}$

Hence, the 95% confidence interval of $T=\sum Y_i$ is