# What am I doing wrong? Joint Density of sum of random variables!

So, I have this question: And, what I have tried so far is I let $z_1 = \frac{X + Y}{2}$ and let $z_2 = \frac{X - Y}{2}$ to use change of variables for this question. However, when I work this out and plug it into my joint distribution of $X$ and $Y$ above, I end up with the $z_2$ disappearing and only having $z_1$ left in my joint distribution of $z_1$ and $z_2$. So, then when I go to find the marginal for just $z_1$ (which is $\frac{X + Y}{2}$ <- what I am looking for), there is no $z_2$ therefore, integrating wrt to $z_2$, I get $z_2$ just come up (integral of 1). This is an issue because the bounds are from 0 to $\inf$. Obviously, I can't have infinite in my marginal density so what am I doing wrong?

Also, for question 5, I work it out similarly except I let $z_2 = X$ but I just end up with the exact same density as $f_{x,y}$ as seen in question 4 - is that normal? Please help, I been doing this question for a long time trying various dummy transformations.

Also, for question 4, I found the marginals of $x$ and $y$ and realized that the independence consequence (of the product being equal to the joint) is satisfied but using convolution, I get the same issue of my integral coming out to infinite! Thank you for all the help!

EDIT: Am I doing my bounds wrong? the bound for z_2 is from 0 to inf or 0 to z_1?

• Convolution, properly understood, is the correct method. But of course you don't use just the marginals--you integrate over the joint distribution. See stats.stackexchange.com/search?q=sum+dependent for many examples. In this simple case a basic approach, derived from the definitions, will work fine: find the chance that $X\le x$ and, simultaneously, $X+Y\le z$. That gives the joint distribution function.
– whuber
Sep 22, 2017 at 21:00
• Sorry, I don't have the best foundation for this kind of stuff, but it looks like in your comment your telling me to find the common ground between X and X+Y but I don't know the best way of going about doing that. Like, I have found the marginals of X and Y thus far and I don't know how I would identify the X+Y in the approach you are mentioning
– ren
Sep 22, 2017 at 21:15
• @whuber Did you mean: P(X <= x) = P(X <= Z - Y) \\ 2 - 2e^(-x) = 2 - 2e^(z-y) or something?
– ren
Sep 22, 2017 at 21:21

First, you can see that $X$ and $Y$ are independent. This is because the joint pdf is: $f_{XY}(x,y)=e^{-(x+y)}=e^{-x} e^{-y}$ (read this for independent random variables if you are not sure).

Also, you can get pdf of $X, Y$ by integration of the joint pdf.

$$f(x)=\int_0^{\infty}e^{-(x+y)}dy=e^{-x}\\f(y)=\int_0^{\infty}e^{-(x+y)}dx=e^{-y}$$

After you know the independence of $X,Y$, to find the distribution $(X+Y)/2$ will be not difficult.

I would like to use moment generating functions for this kind of problems.

Let $Z=\frac{X+Y}{2}$

$$E(e^{tZ})=E[e^{\left( t\frac{X+Y}{2} \right)}]=E(e^{t\frac{X}{2}} e^{t\frac{Y}{2}})$$

$\text{(by independence of X, Y )}$

$$=E(e^{t\frac{X}{2}})E(e^{t\frac{Y}{2}})=\int_{0}^{\infty}e^{t\frac{x}{2}}e^{-x}dx \times \int_{0}^{\infty}e^{t\frac{y}{2}}e^{-y}dy=\frac{1}{(1-\frac{t}{2})^2} , \text{t<2 here}$$

i.e. $$E(e^{tZ})=\frac{1}{(1-\frac{t}{2})^2}$$

$\therefore$ $Z$ has a gamma distribution with $\beta=1/2$ and $\alpha=2$

i.e $f(z)=\frac{1}{\Gamma(1/2)(1/2)^2}ze^{-2z}, 0<z<\infty$

For questions 5. You can use joint distribution of $f_{XY}(x,y)$ and Jacobian, such as let $Z =X+Y$ then $Y=X-Z$. It should not be very difficult

• Thank you so much for the great answer! I just had a follow up question: for question 5, you said I could just use the jacobian but then why could I not just do that for question 4? However, if I do it that way, I end up with a different distribution... So, why can I use it in question 5, but not for question 4 where I would just have the added step of finding the marginal for Z? Thank you.
– ren
Sep 27, 2017 at 22:43
• You can use Jacobian in question 4 , then you better solve question 5 first, then you can solve questin 4 from 5. For question 5 you do this way, Let $Z_1=X, Z_2=X+Y$ and your Jacobian is a $2 \times 2$ determinant. Sep 27, 2017 at 23:15
• But when I use the jacobian, I don't get the gamma distribution, I get e^-z2 * z_2 for 4
– ren
Sep 27, 2017 at 23:34