Probability sum of exponential random variables is less than a value

Question

I've been struggling with the following probability question. It is not a homework question. I just want to know how to do the problem.

Assume that $N, X_{1}, X_{2}, X_{3}, \ldots $are independent random variables where $N$ has as geometric distribution with probability mass function $\Pr\{N = k\} = (1 - p)^{k - 1}p,$ where $k = 1, 2, 3 \ldots, $ and $0 < p < 1$. Moreover, $X_{1}, X_{2}, \ldots$ have an exponential distribution with the probability distribution function

$$f(x) = \begin{cases} \lambda e^{-\lambda x} & \text{ if } x > 0\\ 0, & \text{ if } x \leq 0. \end{cases} $$

Let $Y = X_{1} + X_{2} + \ldots X_{N}$.

• Compute $\Pr\{Y \leq y, N = k\}$ where $y \in (-\infty, \infty)$ and $k = 1, 2, 3, \ldots$

• Find the probability distribution function of $Y$.

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For the first question, I tried to calculate it using small values of $k$ first. For example, suppose $N = 1$. Then we need to find $\Pr\{X_{1} \leq y\}$, which is just $\int_{0}^{y} \lambda e^{-\lambda x} \mathop{dx} = -e^{-\lambda y}$. I couldn't find any patterns, though.

For the second question, I'm aware that the probability distribution for sums of exponential random variables is a gamma distribution, but I'm not sure about how to find it.

Thanks

Answer 1

Note: My analysis assumes $\lambda=1$, but can be readily extended to the $\lambda \neq 1$ case.

We can write $\mathbb{P}(Y \leq y, N=k) = \mathbb{P}(Y \leq y | N=k)\mathbb{P}(N=k)$.

Let's use the fact that the PDF of the sum of two independent random variables is given by the convolution of the the respective PDFs. For $k=1$, Y|N=1 = $X_1 ~\sim exp(\lambda)$. For $k=2$, $Y|N=2=X_1+X_2$ and $f_{Y|N=2}(\cdot) = f_x(\cdot) \star f_x(\cdot)$. The convolution can be computed (by definition) as:

$f_{Y|N=2}(y) = \int_{0}^{y} e^{-x}e^{-(y-x)}dx = ye^{-y} \; \forall \; y \geq 0$.

Extending to $k=3$, the PDF of $Y|N=3 = X_1+X_2+X_3$ can be computed as:

$f_{Y|N=3}(y) = \int_0^y x e^{-x} e^{-(y-x)} dx = \frac{1}{2}y^2 e^{-y} \; \forall \; y \geq 0$.

This can be generalised to (and proved via mathematical induction) to: $f_{Y|N=k}(y) = \frac{1}{(k-1)!}y^{k-1}e^{-y} \; \forall \; y \geq 0$.

Now that we have the PDF, we can compute the CDF as:

$\mathbb{P}(Y \leq y | N=k) = \int_0^y f_{Y|N=k}(x)dx = \int_0^y \frac{1}{(k-1)!}x^{k-1}e^{-x} dx = \frac{1}{(k-1)!}M_k(y)$.

Let's use integration by parts to evaluate the above expression. We have the recursion: $M_k(y) = \int_0^y x^{k-1}e^{-x}dx = -x^{k-1}e^{-x} |_0^y + \int_0^y (k-1)x^{k-2}e^{-x} dx = -y^{k-1}e^{-y} + (k-1)M_{k-1}(y)$. With a bit of algebra one can show that $M_k(y) = (k-1)! - e^{-y} \sum_{j=0}^{k-1} \frac{(k-1)!}{j!} y^j $. Substituting in our previous expression, we get:

$\mathbb{P}(Y \leq y | N=k) = 1 - e^{-y}\sum_{j=0}^{k-1} \frac{y^j}{j!} $.

Finally, the desired joint distribution is given by:

$\mathbb{P}(Y \leq y, N=k) = \left[1 - e^{-y}\sum_{j=0}^{k-1} \frac{y^j}{j!} \right](1-p)^kp$.

You can compute $\mathbb{P}(Y=y) = \sum_{k=1}^\infty \mathbb{P}(Y=y, N=k)$ and plug in the expression computed above. I am not sure if there is a closed form analytical answer.