# Exam P Practice Problem 3

Problem 3a
Suppose that the lifetime (in years) of an electronic device has the following probability density function:

$\displaystyle f(x)=0.0625 \ x \ e^{-0.25 x} \text{where } x>0$

The device is sold with a one-year warranty. The manufacturer is considering offering an extended warranty for an additional one year. What proportion of all devices that are found to be working at the expiration of the regular warranty will be working at the end of the extended warranty?

Problem 3b
Losses under the two policies (policy 1 and policy 2) are independent and each of the losses follows an exponential distribution with mean 2. if the total loss is greater than 3, what is the probability that the loss from policy 1 is greater than 3?

Solution is found below.

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Solution to Problem 3a
The problem is to find the probability that a device will survive two years given that it has already survived one year. Let $X$ be the lifetime. Then we need to find $P(X>2 \lvert X>1)$. Note that the pdf given in the problem is a Gamma distribution with parameters 0.25 and 2. We need to find the following right tail of this distribution:

$\displaystyle \int_t^{\infty} 0.0625 \ x \ e^{-0.25 x} \ dx$

The integral can be evaluated by the method of integration by parts. We can also evaluate using a Poisson distribution. The integral is equivalent to $P(N \le 1)$ where $N$ has the Poisson distribution with parameter $0.25t$ (see Evaluating the Gamma right tail).

\displaystyle \begin{aligned}\int_t^{\infty} 0.0625 \ x \ e^{-0.25 x} \ dx&=e^{-0.25t}+e^{-0.25t} \ (0.25t) \\&=e^{-0.25t} \ (1+0.25t) \end{aligned}

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\displaystyle \begin{aligned}P(X>2 \lvert X>1)&=\frac{e^{-0.5} \ 1.5}{e^{-0.25} \ 1.25} \\&\text{ } \\&=\frac{1.5}{1.25} \ e^{-0.25} \\&\text{ } \\&=0.93456 \end{aligned}

0.4

# Evaluating the Gamma right tail

We present an alternative way of evaluating the following integral, which takes advantage of the connection between Gamma distribution and Poisson process.

$\displaystyle (0) \ \ \ \ \ \ \ \ \ \ \ \ \int_t^{\infty} \frac{\alpha^n}{(n-1)!} \ x^{n-1} \ e^{-\alpha x} \ dx$

The integrand in $(0)$ is the probability density function (pdf) of a Gamma distribution with parameters $\alpha$ (scale parameter) and $n$ (shape parameter) where $\alpha>0$ and $n$ is a positive integer. Thus the integral is the right tail of a Gamma distribution. If the pdf is to model the lifetime of a biological life, the integral is the survival function (i.e. the probability that the life will survice beyond time $t$).

The integral $(0)$ can be evaluated using the method of integration by parts and the following provides the alternative method.

$\displaystyle (1) \ \ \ \ \ \ \ \ \ \ \ \int_t^{\infty} \frac{\alpha^n}{(n-1)!} \ x^{n-1} \ e^{-\alpha x} \ dx=\sum \limits_{j=0}^{n-1}\ \frac{(\alpha t)^j \ e^{-\alpha t}}{j!}$

The goal is not to memorize $(1)$, but rather to understand the thought process behind the formula. The thought process involves identifying the Poisson process that is associated with the Gamma pdf in the integral. The integral is equivalent to a probability calculation in the Poisson process. The right-hand-side of $(1)$ is the probability that there are less than $n$ changes in the Poisson process from time zero to time $t$. The left-hand-side of $(1)$ is the probability that the $n^{th}$ change in the Poisson process occurrs after time $t$. The description of the algorithm is followed by examples and a concluding remark.

The Algorithm

1. From the Gamma pdf in the left-hand-side of $(1)$, pick out the three numbers $\alpha$, $n$ and $t$. The first two are the parameters of the Gamma pdf and the third one is the lower limit of the integral.
2. Consider the Poisson process with parameter $\alpha$. One thing to keep in mind here is that in this Poisson process, the number of changes in a unit time interval is modeled by $\displaystyle P(N_1=j)=\frac{\alpha^j e^{-\alpha}}{j!}$ and the number of changes in a time interval of length $t$ is modeled by $\displaystyle P(N_t=j)=\frac{(\alpha t)^j e^{-\alpha t}}{j!}$
3. The right-hand-side of $(1)$ is the probability that there are less than $n$ changes in the time interval $(0,t)$ (or at most $n-1$ changes).

Example 1

Evaluate $\displaystyle \int_{6}^{\infty} x \ e^{-\frac{x}{3}} \ dx$.

The three parameters in the algorithm are $\alpha=\frac{1}{3}$, $n=2$ and $t=6$. Consider the Poisson process with parameter $\alpha=\frac{1}{3}$. We need to find the probability that there are less than 2 changes in the time interval $(0,6)$. The Poisson probability is $\displaystyle P(N_6=j)=\frac{(2)^j e^{-2}}{j!}$. Thus we have:

\displaystyle \begin{aligned}\int_{6}^{\infty} x \ e^{-\frac{x}{3}} \ dx&=9 \int_{6}^{\infty} \frac{1}{9} \ x^{2-1} \ e^{-\frac{x}{3}} \ dx \\&\text{ } \\&=9 \ \sum \limits_{j=0}^{1}\ \frac{(2)^j \ e^{-2}}{j!} \\&\text{ } \\&=9 \ e^{-2}(1+2)\\&\text{ } \\&=27e^{-2} \end{aligned}

Example 2

Evaluate $\displaystyle \int_{1.5}^{\infty} x^3 \ e^{-2 x} \ dx$.

The three parameters in the algorithm are $\alpha=2$, $n=4$ and $t=1.5$. We are interested in the Poisson process with parameter $\alpha=2$ and in finding the probability that there are less than 4 changes in time interval $(0,1.5)$. The Poisson probability is $\displaystyle P(N_{1.5}=j)=\frac{(3)^j e^{-3}}{j!}$. The integral, after adjusting for a multiplicative constant, is the probability that the fourth change occurs after time 1.5.

\displaystyle \begin{aligned}\int_{1.5}^{\infty} x^3 \ e^{-2x} \ dx&=\frac{6}{16} \int_{1.5}^{\infty} \frac{2^4}{3!} \ x^{4-1} \ e^{-2x} \ dx \\&\text{ } \\&=\frac{3}{8} \ \sum \limits_{j=0}^{3}\ \frac{(3)^j \ e^{-3}}{j!} \\&\text{ } \\&=\frac{3}{8} \ e^{-3}(1+\frac{3}{1!}+\frac{3^2}{2!}+\frac{3^3}{3!})\\&\text{ } \\&=\frac{39}{8} e^{-3} \end{aligned}

Remark
The reason that the summation in the right-hand-side of $(1)$ works as an answer to the integral on the left-hand-side is that the summation is how the Gamma pdf is derived (where the shape parameter is a positive integer). Let $X$ be the time until the occurrence of the $n^{th}$ change in the Poisson process that has parameter $\alpha$. Then the right-hand-side of $(1)$ is $S_X(t)=P(X>t)$. The distribution function is then $F_X(t)=1-S_X(t)$. Taking derivative of $F_X(t)=1-S_X(t)$ produces the integrand in the left-hand-side of $(1)$. Thus the alternative approach indicated here is simply going back to the root of the Gamma pdf.