Find the density of the sum of two jointly distributed rv's

[Eclair_de_XII]

Homework Statement

"Given that the joint distribution of $X$ and $Y$ is $f(x,y)=\frac{1}{2}(x+y)e^{-(x+y)},\text { for } x,y>0$ and $0$ otherwise, find the distribution of $Z=X+Y$."

Homework Equations

$f_Z(z)=\int_{\mathbb{R}}f(x,z-x)dx$
$P(Z<z)=F_Z(z)=P(X+Y<z)=P(Y<z-X)=\int_{\mathbb{R}} \int_{-\infty}^{z-x}f(x,y)dydx$

The Attempt at a Solution

$f_Z(z)=\int_0^\infty \frac{1}{2}[x+(z-x)]e^{-[x+(z-x)]}dx=\frac{1}{2} \int_0^\infty ze^{-z}dx=\frac{1}{2}ze^{-z}x|_0^\infty$

The problem is that the density function seems to always tend towards infinity no matter which equation I use...

 

[Charles Link]

It's probably not that simple, but why isn't $f(z)=\frac{1}{2}z \, e^{-z}$ for z>0 ?

 

[StoneTemplePython]

my motto is start simple and build. I'd start by integrating out / marginalizing $Y$ to get the marginal of $X$ -- you can either re-run the process or use symmetry to then get the marginal of $Y$. Then confirm that your calculated distribution for $X$ is a bonafide random variable-- i.e. that the density integrates to one.

Then convolve these two. The convolution (sum) of two bonafide random variables always produces an actual random variable.

$f_Z(z)=\int_{\mathbb{R}}f(x,z-x)dx$
$P(Z<z)=F_Z(z)=P(X+Y<z)=P(Y<z-X)=\int_{\mathbb{R}} \int_{-\infty}^{z-x}f(x,y)dydx$

You need to be a lot more careful here. We've been through this before -- while you can get away with sloppy use of strictness of inequalities for purely continuous random variables, at some point this is going to cause immense problems and you'll be blind sided by it. The convention for a CDF is $P(Z\leq z)=F_Z(z)$ -- I highly doubt your book by Grimmett breaks this convention.

by the way in terms of pattern recognition, it should jump out at you that you're dealing with something exponential random variable-ish. For a lot of uses they are the nicest behaved random variables. I'd probably say to just do the convolution directly, but since this is heavily using the exponential, you may want to try using the Moment Generating Function to do the convolution.

 

[Ray Vickson]

Homework Statement

"Given that the joint distribution of $X$ and $Y$ is $f(x,y)=\frac{1}{2}(x+y)e^{-(x+y)},\text { for } x,y>0$ and $0$ otherwise, find the distribution of $Z=X+Y$."

Homework Equations

$f_Z(z)=\int_{\mathbb{R}}f(x,z-x)dx$
$P(Z<z)=F_Z(z)=P(X+Y<z)=P(Y<z-X)=\int_{\mathbb{R}} \int_{-\infty}^{z-x}f(x,y)dydx$

The Attempt at a Solution

$f_Z(z)=\int_0^\infty \frac{1}{2}[x+(z-x)]e^{-[x+(z-x)]}dx=\frac{1}{2} \int_0^\infty ze^{-z}dx=\frac{1}{2}ze^{-z}x|_0^\infty$

The problem is that the density function seems to always tends towards infinity no matter which equation I use...

You are correct that $f_Z(z) = \int_R f(x,z-x) \, dx,$ and with the given form of $f(x,y)$ this is about as simple as you can get. Remember also that the integral does not go from $x = -\infty$ to $x = +\infty$, because of the way the problem was specified; you need to figure out the actual $x$-range for a given value of $z > 0.$

 

[StoneTemplePython]

It's probably not that simple, but why isn't $f(z)=\frac{1}{2}z \, e^{-z}$ for z>0 ?

I always do the expected values first. If you use this, then
$E\big[Z\big] = \int_{0}^\infty z \big(\frac{1}{2}z \, e^{-z}\big) = 1$

but
$E\big[Z\big] = E\big[X + Y\big] = E\big[X\big] + E\big[Y\big] = 2 \cdot E\big[X\big] = 3 \gt 1$

- - - -
a more direct explanation is that $f(z)=\frac{1}{2}z \, e^{-z}$ integrates to one half, not one, so it's not a valid distribution. You could rescale by 2 to make it integrate to one, but you still have a problem with the expected value.

 

[Ray Vickson]

I always do the expected values first. If you use this, then
$E\big[Z\big] = \int_{0}^\infty z \big(\frac{1}{2}z \, e^{-z}\big) = 1$

but
$E\big[Z\big] = E\big[X + Y\big] = E\big[X\big] + E\big[Y\big] = 2 \cdot E\big[X\big] = 3 \gt 1$

- - - -
a more direct explanation is that $f(z)=\frac{1}{2}z \, e^{-z}$ integrates to one half, not one, so it's not a valid distribution. You could rescale by 2 to make it integrate to one, but you still have a problem with the expected value.

Another way would be to compute the Laplace transform of the density of $X+Y$, which is just
$$\int_{x=0}^\infty \, \int_{y=0}^\infty f(x,y) e^{-s(x+y)} \, dx \, dy.$$

 

[Eclair_de_XII]

Remember also that the integral does not go from $x = -\infty$ to $x = +\infty$, because of the way the problem was specified; you need to figure out the actual $x$-range for a given value of $z>0$.

So I drew a picture (it still surprises me how helpful this is outside of physics class)...

Basically, I wanted to find the area shaded pink. I set my bounds...

$0\leq y \leq z - x$
$0 \leq x \leq z$

And I want to find $P(X+Y\leq z)=P(Y\leq z - X)$. So...

$P(Y\leq z - X)=\int_0^z \int_0^{z-x} \frac{1}{2}(x+y)e^{-(x+y)}dydx$

Some substitution...

$u=(x+y)$, $du=dy$
$z-x\mapsto z$, $0\mapsto x$

So we have:

$P(Y\leq z - X)=\frac{1}{2} \int_0^z \int_x^z ue^{-u}dudx=\frac{1}{2} \int_0^z[-ue^{-u}-e^{-u}]|_x^z\\ =\frac{1}{2} \int_0^z [(-ze^{-z}-e^{-z})-(-xe^{-x}-e^{-x})]dx\\ =\frac{1}{2}(-ze^{-z}-e^{-z})x|_0^z+\frac{1}{2}\int_0^z (xe^{-x}+e^{-x})dx\\ =\frac{1}{2}(-z^2e^{-z}-ze^{-z})+\frac{1}{2}(-xe^{-x})|_0^z\\ =\frac{1}{2}(-z^2e^{-z}-2ze^{-z})=F_Z(z)$

Something's not right here. $F_Z(z)$ isn't supposed to be negative. I think I forgot to find the marginal of $X$ and integrate that along with the distribution function. On the other hand, I could set the upper bound for $x$ as $z$, and use my original integrand to get the answer in the back of the book. But I cannot be sure that it's enough.

 

[Orodruin]

$$
\frac{d}{dx} (-x e^{-x}) = - e^x + x e^{-x} \neq x e^{-x} + e^{-x}
$$

 

[Ray Vickson]

So I drew a picture (it still surprises me how helpful this is outside of physics class)...

View attachment 234165

Basically, I wanted to find the area shaded pink. I set my bounds...

$0\leq y \leq z - x$
$0 \leq x \leq z$

And I want to find $P(X+Y\leq z)=P(Y\leq z - X)$. So...

$P(Y\leq z - X)=\int_0^z \int_0^{z-x} \frac{1}{2}(x+y)e^{-(x+y)}dydx$

.........

Something's not right here. $F_Z(z)$ isn't supposed to be negative. I think I forgot to find the marginal of $X$ and integrate that along with the distribution function. On the other hand, I could set the upper bound for $x$ as $z$, and use my original integrand to get the answer in the back of the book. But I cannot be sure that it's enough.

You are doing way too much work, as you already had the correct expression back in post #1:
$$f_Z(z) = \int_R f(x,z-x) \, dx.$$ However, you forgot that $f=0$ outside the region $\{ x \geq 0, z-x \geq 0 \}$, and inside this region the integrand $f(x,z-x) =$ a very simple function of $z$ alone.

As to where your error occurred: you computed $\int_0^z (x e^{-x} + e^{-x} ) \, dx$ incorrectly. Use $\int_0^z e^{-x} \, dx = 1-e^{-z}$ and $\int_0^z x e^{-x} \, dx = 1 - (z+1) e^{-z}.$

 

[Eclair_de_XII]

Use $\int_0^z e^{-x} \, dx = 1-e^{-z}$ and $\int_0^z xe^{−x} \, dx=1−(z+1)e^{−z}$.

$P(Y\leq z - X)=\frac{1}{2}(-z^2e^{-z}-ze^{-z})+\frac{1}{2}(1-e^{-z}+1-(z+1)e^{-z})\\ =\frac{1}{2}(-z^2e^{-z}-2ze^{-z}+2-2e^{-z})=F_Z(z)$

$\frac{d}{dz}F_Z(z)=\frac{1}{2}[(-2ze^{-z}+z^2e^{-z})-(2e^{-z}-2ze^{-z})+2e^{-z})=\frac{1}{2}z^2e^{-z}=f_Z(z)$

Okay, thank you everyone for your help.

 

[StoneTemplePython]

So we end up with $Z$ is Erlang of order 3 with $\lambda = 1$. Given the exponential underneath I wonder if there is an 'easy' way, other than pattern recognition, of going from this

Another way would be to compute the Laplace transform of the density of $X+Y$, which is just
$$\int_{x=0}^\infty \, \int_{y=0}^\infty f(x,y) e^{-s(x+y)} \, dx \, dy.$$

(or with a negative sign in front of s, the MGF) to the final PDF for Z...

 

[Ray Vickson]

$P(Y\leq z - X)=\frac{1}{2}(-z^2e^{-z}-ze^{-z})+\frac{1}{2}(1-e^{-z}+1-(z+1)e^{-z})\\ =\frac{1}{2}(-z^2e^{-z}-2ze^{-z}+2-2e^{-z})=F_Z(z)$

$\frac{d}{dz}F_Z(z)=\frac{1}{2}[(-2ze^{-z}+z^2e^{-z})-(2e^{-z}-2ze^{-z})+2e^{-z})=\frac{1}{2}z^2e^{-z}=f_Z(z)$

Okay, thank you everyone for your help.

Even easier and faster (following your post #1):
$$f_Z(z) = \int_{x=0}^z \underbrace{f(x,z-x)}_{(1/2) z e^{-z}} \, dx = (1/2) z^2 e^{-z}.$$