Suppose that $X$ and $Y$ are independent random variables that follow an exponential distribution with mean $1$ and let $W = Y - X$.

(i) Using the cumulative distribution function (cdf), find the probability density function (pdf) of $W$, $f_W(w)$ for $w > 0$. (No need to compute $w \leq 0$).

(ii) Determine the moment generating function (mgf) of $W$ by finding the mgfs of $Y$ and $-X$. Justify where the mgf is defined.

I'm not quite sure how to do this problem but here is what I did:


Suppose $X, Y$ have means $\lambda_1$ and $\lambda_2$. If $W = Y - X$ then (I think)

$$P(W \leq w) = \int_0^\infty \int_{0}^{x+z} \lambda_1 e^{-\lambda_1 x} \lambda_2 e^{-\lambda_2 y}\,dy \ \ dx$$

If we substitute $\lambda_1 = \lambda_2 = 1$ then

$$P(W \leq w) = \int_0^\infty \int_{0}^{x+z} e^{-x - y}\,dy \ \ dx$$


The moment generating function for an exponential distribution is $M(t) = \frac{1}{1- \theta t}$ where $\theta > 0$ and $t < \frac{1}{\theta}$.

If the mean $\theta = 1$ then we know that for $Y$

$M_Y(t) = \frac{1}{1-t}$

As for $-X$, I'm not sure.

I'm not sure about the formula in (i) or how to derive it, and I'm confused on finding the mgf of $-X$ in (ii). But in any case I am stuck. Any assistance will be much appreciated.


1 Answer 1


For the formula in (i), $$ \mathbb{P}(W \leq w) = \mathbb{P}(f(X,Y) \in ]-\infty,w]) = \mathbb{P}((X,Y) \in f^{-1}(]-\infty,w]))$$ where $f(x,y) = y-x$. Since $X \sim \mathcal{E}(\lambda_1)$ and $Y \sim \mathcal{E}(\lambda_2)$ are independent and admit both density, denoted $f_X$ and $f_Y$, the couple $(X,Y)$ admites the density $f_{(X,Y)}(x,y) = f_X(x)f_Y(y)$ and thus \begin{align} \mathbb{P}((X,Y) \in f^{-1}(]-\infty,w])) &= \int_{f^{-1}(]-\infty,w])} \lambda_1e^{-\lambda_1x}\lambda_2e^{-\lambda_2y}1_{x >0}1_{y>0}dxdy\\ &= \int_{\mathbb{R}^2} 1_{y-x \leq w} \lambda_1e^{-\lambda_1x}\lambda_2e^{-\lambda_2y}1_{x >0}1_{y>0}dxdy\\ &= \int_0^{\infty} \int_0^{\infty} 1_{y \leq x+w}\lambda_1e^{-\lambda_1x}\lambda_2e^{-\lambda_2y}dxdy \\ &= \int_0^{\infty} \int_0^{\infty} 1_{y \leq x+w}\lambda_1e^{-\lambda_1x}\lambda_2e^{-\lambda_2y}dydx \\ &= \int_0^{\infty} \int_0^{\infty} 1_{y \leq x+w}1_{x+w \geq 0}\lambda_1e^{-\lambda_1x}\lambda_2e^{-\lambda_2y}dydx \\ \end{align} and now you can adapt the rest of the computations depending on the sign of $w$.

For (ii), for all $t \in ]-1,1[$, $$ M_{-X}(t) = \mathbb{E}[e^{-tX}] = M_X(-t) = \frac{1}{1+t}.$$

  • $\begingroup$ Unfortunately, I am relatively new to distributions and thus not familiar with these theorems.. $\endgroup$
    – user902292
    Mar 18, 2021 at 21:05
  • $\begingroup$ Which part are you unfamiliar with? $\endgroup$ Mar 18, 2021 at 21:07
  • $\begingroup$ Part (i). I have no idea what Lebesgue measures are or the Fubini-Tonelli theorem etc. I don't think I am supposed to use these for this problem. $\endgroup$
    – user902292
    Mar 18, 2021 at 21:08
  • 1
    $\begingroup$ Ok let me rewrite this into a probabilistic language $\endgroup$ Mar 18, 2021 at 21:09
  • 1
    $\begingroup$ Yes it is, note that the fourth equality holds because $X$ and $Y$ are independent. $\endgroup$ Mar 18, 2021 at 21:41

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