# Let $X_1$ and $X_2$ be two independent exponential random variables with $\lambda_1 \neq \lambda_2$ as parameters

Let $$X_1$$ and $$X_2$$ be two independent exponential random variables with $$\lambda_1 \neq \lambda_2$$ as parameters. I want to find the density function of $$Z = aX_1 + bX_2$$ with $$a,b > 0$$.

This is what I have done so far:

$$f_Z(z) = \int_{-\infty}^{+\infty}f_{X_1}(z-x_2)f_{X_2}(x_2) dx_2$$

$$= \int_0^z a\lambda_1 e^{-a\lambda_1(z-x_2)}b\lambda_2 e^{-b\lambda_2x_2}dx_2$$ and from here calculations. I just wanted to make sure that this procedure is the right one.

## 1 Answer

There's one mistake you're making. Bear in mind $$bX_2\sim\operatorname{Exp}(\lambda_2/b)$$, so as to scale the mean from $$1/\lambda_2$$ to $$b/\lambda_2$$. Thus each appearance of $$a,\,b$$ in your integral should be replaced with $$1/a,\,1/b$$. Apart from that, though, you're on the right track.

• I could sense there was something wrong. Thank you! – qcc101 Nov 5 '18 at 20:04
• @qcc101 No problem. You could also have spotted it by using dimensional analysis. – J.G. Nov 5 '18 at 20:08