0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – André Nicolas Apr 9 '16 at 7:26
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.