Joint density of two exponential random variables

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 $$f_{Z, W}$$

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

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 $$g^{-1}$$.

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 $$\int_0^\infty \int_{-\infty}^w e^{z-2w} dzdw = 1$$

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 $$Z$$: $$f_Z(t) = \begin{cases} \frac{\lambda \mu e^{\frac{\mu t}{2}}}{2 \lambda +\mu },& t<0\\ \frac{\lambda \mu e^{-\lambda t}}{2 \lambda +\mu },& t>0 \end{cases}.$$

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.

• 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 Jan 11, 2020 at 21:09
• No of course not. $X, X-2Y$ are not independent because intuitively $X-2Y$ depends on $X$. Jan 11, 2020 at 21:42