# Intuitive explanation of distribution of ratio of independent random variables

Case 1: I have two independent exponentially distributed random variables $X$ and $Y$. Intuitively, it makes sense that the sum of those variables is essentially exponentially distributed, but is that correct?

Case 2: I have a uniform random variable $X$ in $(0.75, 1.25)$ (i.e. with mean 1) and an independent exponential random variable $Y$ with parameter $m$. Intuitively, it makes sense that the ratio $\frac{Y}{X}$ is essentially exponentially distributed, but is that correct?

• The sum of two independent exponentials is not exponential. For the second, I have not written out details, but am pretty sure it will not be exponential. – André Nicolas Apr 9 '16 at 7:26

Case 2: We are essentially interested in $$\mathbb{P}\left(\frac{Y}{X} < t\right) = \mathbb{P}(Y < tX) \quad(*).$$ Since $X$ and $Y$ are assumed to be independent we have the following expression for the joint density \begin{align}f_{X,Y}(x,y) &= \frac{1}{2} I_{x \in [0.75,1.25]} m e^{-my}I_{y \geq 0}. \end{align} Hence to find $(*)$ you have to compute the following integral \begin{align} &\int_{0.75}^{1.25} \int_{0}^{tx} f_{X,Y}(x,y) dx dy \\ \end{align} As you will see, the ratio $Y/X$ is not exponentially distributed.