PDF of sum exponential random variable multiplied by constant.

Question

I am trying to find the PDF of $\gamma_m$ inthe following expression but getting stuck due to presence of summation and constant.

$\gamma_m = \sum_{i=1}^{R}k_i\cdot X_i$-----(1)

where, $X_i$ is i.i.d. exponential random variable and $k_i$ is the constant.

Any help pls....

Answer 1

Let $$ \gamma_m = k\sum_{i=1}^n X_i$$ where $X_i \sim \exp (\lambda) $ are i.i.d. Consider the moment generating function $$ M_{\gamma_m}(t):= \mathbb{E} e^{t \gamma_m}.$$ Then \begin{align*} M_{\gamma_m}(t) &= \mathbb{E} e^{tk \sum_{i=1}^n X_i} \\ &= \mathbb{E} \prod_{i=1}^m e^{tk X_i}. \end{align*} Since $X_i$ are i.i.d., we are able to swap the product and the expected value. If $M(t) = \mathbb{E} e^{t X_i}$ then it follows $$M_{\gamma_m}(t) = \prod_{i=1}^m \mathbb{E}e^{t kX_i} = (M(kt))^m. $$ It is well known that the moment generating function of an exponential distribution is $$M(t) = \frac{\lambda}{\lambda-t} \qquad (t<\lambda). $$ Thus, $$M_{\gamma_m}(t) = \bigg ( \frac{\lambda}{\lambda-kt}\bigg )^m\qquad (t<\frac \lambda k). $$ From inspection, this is the moment generating function of a $\Gamma(m,\lambda/k)$. Thus, $\gamma_m \sim \Gamma(m,\lambda/k)$.

Remarks: 1. If you don't assume $k_i = k$ then $$M_{\gamma_m}(t) = \prod_{i=1}^m\frac{\lambda}{\lambda-k_it}.$$ It is not altogether clear what distribution this corresponds to.

2. Some people might argue that this calculation is tautological since a $\Gamma(m,\lambda)$ r.v. is often defined to be the sum of $m$ i.i.d $\exp(\lambda)$ r.v's.