I am trying to find PDF that involves sum of two i.i.d exponential random variables multiplied by a constant i.e.,
$$\gamma = c_1 X_1 + c_2X_2 \tag 1$$
where both $c_1,c_2$ are constant and $X_1,X_2$ are exponential random variable. This is what I had tried:
Let us represent (1) as: $\gamma = \gamma_1 + \gamma_2$. Therefore PDF of $\gamma_1$, $\gamma_2$ can be written as $f_{\gamma_1}(x_1) = \frac{1}{c_1\sigma_1^2}\text{exp}\left(\frac{-x_1}{c_1\sigma_1^2}\right)$ and $f_{\gamma_2}(x_2) = \frac{1}{c_2\sigma_2^2}\text{exp}\left(\frac{-x_2}{c_2\sigma_2^2}\right)$ respectively.
Next using convolution we have
$$f_{\gamma}(x) = \int_{-\infty}^{{\infty}}f_{\gamma_1}(x_1)f_{\gamma_1}(x-x_1) \,\text{d}x$$
On solving this, finally we get
$$f_{\gamma}(x) = \frac{1}{c_1\sigma^2_1} \text{exp}\left(\frac{x_1}{c_2\sigma^2_2}-\frac{x_1}{c_1\sigma^2_1}\right)\tag 2$$
My query is that whether the PDF obtained in $(2)$ correct?
Any help in this regard will be highly appreciated.
The question says $X_1$ and $X_2$ are independent identically distributed random variables.
So,
$f_{X_1}(x_1) = \lambda e ^{- \lambda x_1}, \ f_{X_2}(x_2) = \lambda e ^{- \lambda x_2}$
We need to find distribution of $\gamma = c_1X_1 + c_2 X_2$
Assuming $c_1, c_2$ are positive and $c_1 \ne c_2$, using convolution,
$\displaystyle f_\gamma(\gamma) = \cfrac{1}{c_2} \int_0^{\gamma / c_1} f_{X_1} (x_1) \cdot f_{X_2} \left(\frac{\gamma - c_1x_1}{c_2} \right) \ dx_1$
Solving you should get,
$ \displaystyle f_{\gamma} (\gamma) = \frac{\lambda}{c_2 - c_1} \left[e^{ - \frac{\lambda}{c_2} \gamma} - e^{ - \frac{\lambda}{c_1} \gamma}\right], \ 0 \leq \gamma \leq \infty$