1
$\begingroup$

I am having trouble understanding how to find the pdf $f_Z(z)$ when $f_{X,Y}(x,y) = e^{-x-y}, x,y \space \epsilon(0,\infty)$ where $Z = X+Y$

My approach is that

$$x = x, y = z-x$$

so using transformation

$$f_z(z)=f_{X,Y}(x,z-x)|J|$$

where J is the Jacobian.

From this I calculated

$$f_Z(z) = e^{-z}$$

but it does not seem to be the correct answer...

What did I do wrong?

The Jacobian that I calculated was

$$ \left | \begin{matrix} \frac{\partial x}{\partial x} =1 & \frac{\partial x}{\partial z} =0 \\ \frac{\partial y}{\partial x} =1 & \frac{\partial y}{\partial z} =-1 \\ \end{matrix} \right | = 1 $$

and I am not confident that this is right.

$\endgroup$
  • $\begingroup$ There are many solutions to $x+y=z$, you need to integrate over them, so what you seek is $f_Z(z) = \int_{0}^\infty f_{X,Y}(x,z-x) dx$. Ususally in order to not make any mistakes, I use the cdf instead because it is always a probability and not a density. $\endgroup$ – P. Quinton Feb 17 at 7:30
  • $\begingroup$ In your formula, it should be $f_{X,Z}(x,z)$ on the left hand side, not $f_Z$. Once you get the joint pdf for $X,Z$ you marginalize to get $f_Z$. This is basically equivalent to P. Quinton's comment. $\endgroup$ – GReyes Feb 17 at 7:36
  • $\begingroup$ Thank you for your help! I will see where I can get. $\endgroup$ – hyg17 Feb 17 at 7:38
  • $\begingroup$ Also, observe that $f_{X,Y}(x,z-x)=e^z$ only if $z-x>0$! $\endgroup$ – GReyes Feb 17 at 7:46
  • $\begingroup$ You could choose for an alternative: from $f_{X,Y}(x,y)=e^{-x-y}$ it follows (almost) immediately that $X,Y$ are iid with standard exponential distribution. Based on that you can find $1-F(z)=P(Z>z)$ and take the derivative of $F$. Personally I am not fond of Jacobians and always try to avoid them. $\endgroup$ – drhab Feb 17 at 8:04
1
$\begingroup$

Approach 1: By the joint density of $(X, Y)$, it can be seen that $X, Y$ i.i.d. $\sim \text{Exp(1)}$, therefore $Z = X + Y$ has $\Gamma(2, 1)$ distribution (this is a property of exponential distribution, see this link).

Approach 2: Calculate the distribution function of $Z$ directly, for $z > 0$: \begin{align} & P[Z \leq z] = P[X + Y \leq z] \\ = & \int_0^z\int_0^{z - x} f_{X, Y}(x, y)dy dx \\ = & \int_0^z \int_0^{z - x}e^{-x - y}dy dx \\ = & \int_0^z e^{-x}(1 - e^{-(z - x)})dx \\ = & \int_0^ze^{-x}dx - \int_0^z e^{-z}dx \\ = & 1 - e^{-z} - ze^{-z}. \end{align} Therefore, $$f_Z(z) = \frac{dP[Z \leq z]}{dz} = e^{-z} - (e^{-z} - ze^{-z}) = ze^{-z}$$, which is the density of $\Gamma(2, 1)$ distribution.

Approach 3: This is your attempt. You may not use $x$ in both the pre-transformation and post-transformation versions. It's better to write the transformation as $$\begin{cases} W = X \\ Z = X + Y. \end{cases}$$ The Jacobian then becomes \begin{align}\frac{\partial(x, y)}{\partial(w, z)} = & \det\begin{pmatrix}\partial x / \partial w & \partial x / \partial z \\ \partial y / \partial w & \partial y / \partial z \end{pmatrix} \\ = & \det\begin{pmatrix} 1 & 0 \\ -1 & 1 \end{pmatrix} = 1. \end{align} Hence the joint density of $(W, Z)$ is $f_{W, Z}(w, z) = e^{-w - (z - w)} \times 1 = e^{-z}, 0 < w < z < \infty$. Finally, do marginalization (pay attention to the integration range of $W$, which probably is the reason you didn't get the correct answer): \begin{align} f_Z(z) = \int_0^z e^{-z} dw = ze^{-z}, 0 < z < \infty. \end{align}

While Approach 1 might be the most expedient, as a general approach, personally I prefer Approach 2 to Approach 3, which is usually more straightforward.

$\endgroup$
  • $\begingroup$ +1 Remark on approach 2: I would rather go for $P(Z>z)$. Makes calculations more easy. $\endgroup$ – drhab Feb 17 at 8:06
  • $\begingroup$ @drhab In my opinion, go for $P(Z \leq z)$ or $P(Z > z)$ are equivalently convenient in this problem. Calculating $P(Z > z)$ is not significantly easier than $P(Z \leq z)$ (maybe slightly easier as you work with one term throughout). $\endgroup$ – Zhanxiong Feb 17 at 8:10
  • $\begingroup$ My preference is actually based on a slightly different approach in calculation. For $z>0$ we have:$$P\left(Z>z\right)=\int_{0}^{z}P\left(Z>z\mid X=x\right)f_{X}\left(x\right)dx+\int_{z}^{\infty}P\left(Z>z\mid X=x\right)f_{X}\left(x\right)dx=$$$$\int_{0}^{z}e^{-z}dx+\int_{z}^{\infty}e^{-x}dx=ze^{-z}+e^{-z}$$ Negative derivative is $ze^{-z}$. $\endgroup$ – drhab Feb 17 at 8:28
  • $\begingroup$ Thank you so much! It's just that the fact that $f(z)$ is the marginal does not click with me very well, so I will approach this problem using probability then take the derivative. Thank you for clarifying my approach, though. That was awesome. $\endgroup$ – hyg17 Feb 17 at 8:38

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.