0
$\begingroup$

For independent random variables X ∼ Exp(1) and Y ∼ Exp(2), find the density of (Z, W) = (X-2Y, X).

My approach:

Since for any exponential distribution with parameter $\lambda$ the function is $f(x) = \lambda e^{-\lambda x}$

$f_X(x) = e^{-x}$

$f_Y(y) = 2e^{-2y}$

Therefore the joint density function is: $$ f_{X, Y}(x, y) =f_X(x) f_Y(y) = \begin{cases} 2e^{-x-2y} \ & \mbox{ if } x \geq 0, y \geq 0; \\ 0 \ & \mbox{ elsewhere}. \end{cases} $$

However I don't know how to use this to calculate $f_{Z, W}$

$\endgroup$
2
  • $\begingroup$ Don't you know any theorems that allow you to transform random vectors? $\endgroup$ Jan 11, 2020 at 20:54
  • $\begingroup$ Any suggestions? $\endgroup$ Jan 11, 2020 at 21:12

3 Answers 3

2
$\begingroup$

The map $g:(x,y) \mapsto (x-2y,x)$ is a differentiable and invertible function between $(0,\infty)\times (0,\infty)$ and $R=\{(z,w) | z< w \text{ and } w>0\}$, so first of all we get that the support for $(Z,W)=(X-2Y,X)$ must be $R$.

The transformation theorem for probability densities states that:

$$f_{Z,W}(z,w) = f_{X,Y}(g^{-1}(z,w)) |det(\frac{dg^{-1}}{d(z,w)}(z,w))|,$$ where $\frac{dg^{-1}}{d(z,w)}(z,w)$ is the jacobian of $g^{-1}$.

(see https://en.wikipedia.org/wiki/Probability_density_function#Vector_to_vector)

We first compute $g^{-1}(z,w)= (w,\frac{w-z}{2})$ and the jacobian $$ \frac{dg^{-1}}{d(z,w)}(z,w) = \begin{pmatrix}0 & 1 \\ -\frac12 & \frac12 \end{pmatrix},$$ which has determinant $\frac12$ for all $z,w$. We now plugin, and get $$ f_{Z,W}(z,w) = \frac12 f_{X,Y} ((w,\frac{w-z}{2})) = e^{-w}e^{-2\frac{w-z}{2}}=e^{z-2w}.$$ for all $(z,w) \in \{(z,w) | z< w \text{ and } w>0\}$. Just to verify, that this is in fact a valid density we compute $$ \int_0^\infty \int_{-\infty}^w e^{z-2w} dzdw = 1$$

$\endgroup$
0
$\begingroup$

Note that while $X$ and $Y$ are nonnegative, $X-2Y$ may take negative values. So we need to treat these cases separately. For $t<0$, we have \begin{align} F_Z(t) &= \mathbb P(Z\leqslant t)\\ &= \mathbb P(2Y\geqslant X-t)\\ &= \int_0^\infty\int_{x-t}^\infty \lambda e^{-\lambda x}\frac\mu 2 e^{-\frac\mu 2y}\ \mathsf dy\ \mathsf dx\\ &= \frac{2 \lambda e^{\frac{\mu t}{2}}}{2 \lambda +\mu }. \end{align} For $t>0$, we have \begin{align} F_Z(t) &= \mathbb P(Z\leqslant t)\\ &= 1 - \mathbb P(2Y\geqslant X-t)\\ &= \int_t^\infty \int_0^{x-t} \lambda e^{-\lambda x}\frac\mu 2 e^{-\frac\mu 2y}\ \mathsf dy\ \mathsf dx\\ &= 1-\frac{\mu e^{-\lambda t}}{2 \lambda +\mu }. \end{align} Differentiating yields the density of $Z$: $$ f_Z(t) = \begin{cases} \frac{\lambda \mu e^{\frac{\mu t}{2}}}{2 \lambda +\mu },& t<0\\ \frac{\lambda \mu e^{-\lambda t}}{2 \lambda +\mu },& t>0 \end{cases}. $$

$\endgroup$
-3
$\begingroup$

If I'm not wrong, you've formed the joint density function under the assumption that the random variables $X$ and $Y$ are independent.

In the formula for $f_{X,Y}(x,y)$, let us put $X-2Y$ in place of $X$, $X$ in place of $Y$, and let us also put $x-2y$ in place of $x$ and $x$ in place of $y$ to obtain $$ f_{Z, W}(z, w) = 2e^{- (x-2y) - x } = 2 e^{2y-2x} $$ if $x-2y \geq 0$ and $x\geq 0$, that is, if $x \geq \max(2y, 0)$, and $0$ otherwise.

However, I'm not completely sure if my solution is correct.

$\endgroup$
2
  • $\begingroup$ Sorry for confusion, yes X and Y are independent but I think that independence of X and Y doesn't imply independence of X-2Y and X $\endgroup$ Jan 11, 2020 at 21:09
  • 1
    $\begingroup$ No of course not. $X, X-2Y$ are not independent because intuitively $X-2Y$ depends on $X$. $\endgroup$ Jan 11, 2020 at 21:42

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.