Random variable independence

[sneaky666]

Homework Statement

Let X~Bernoulli(θ) and Y~Geometric(θ), with X and Y independent. Let Z=X+Y. What is the probability function of Z?

Homework Equations

The Attempt at a Solution

I am getting
PX(1) = θ
PX(0) = 1-θ
PX(x) = 0 otherwise
pY(y) = θ(1-θ)^y for y >= 0
pY(y) = 0 otherwise


not sure where to go from here...

 

[jbunniii]

Do you know this result? The probability distribution of $Z$ is the convolution of the probability distributions of $X$ and $Y$.

 

[sneaky666]

no i am not sure, which is why i need help.

 

[jbunniii]

There's a quick proof of the convolution result at the top of this PDF file:

http://www.dartmouth.edu/~chance/teaching_aids/books_articles/probability_book/Chapter7.pdf

That should get you started. If you get stuck, post what you have and I'll try to help.

 

[sneaky666]

i did that research and the only thing i could come up with is

PX(X=1) = θ
PX(X=0) = 1-θ
PX(X=x) = 0 otherwise
PY(Y=y>=0) = θ(1-θ)^y
PY(Y=y) = 0 otherwise

so (X=k) and (Y=z-k) since Z = X+Y

PZ(Z=z)=

summation from -inf to inf
θ^2 * (1-θ)^(z-1)
if x=1,y=1

summation from -inf to inf
θ * (1-θ)^(z+1)
if x=0,y=0

0
otherwise


Is this right?

 

[jbunniii]

I don't think that's quite right. I am going to introduce some notation to make it easier to express the functions:

Kronecker delta function

$$\delta(k) = \begin{cases} 1, & k = 0 \\ 0, & \textrm{otherwise} \end{cases}$$

Unit step function

$$u(k) = \begin{cases} 1, & k \geq 0 \\ 0, & \textrm{otherwise} \end{cases}$$

Bernoulli distribution

$$b(k) = (1-\theta)\delta(k) + \theta \delta(k-1)$$

Geometric distribution

$$g(k) = \theta (1-\theta)^k u(k)$$

Distribution of sum of independent bernoulli and geometric

$$\begin{align*} s(k) &= \sum_{m=-\infty}^{\infty} g(m) b(k-m) \\ &= \sum_{m = 0}^{\infty} \theta(1 - \theta)^m [(1-\theta)\delta(k-m) + \theta \delta(k-m-1)] \end{align*}$$

You can now consider three cases:

1) $k < 0$: the sum is zero
2) $k = 0$: only one of the two $\delta$ functions is nonzero for some $m \geq 0$
3) $k > 0$: both $\delta$ functions are nonzero for some $m \geq 0$

 

[sneaky666]

ok i see, but i looked on wikipedia and for the bounoulli dist. and geometric dist. i thought it was

PX(X=1) = θ
PX(X=0) = 1-θ
PX(X=x) = 0 otherwise
PY(Y=y>=0) = θ(1-θ)^y
PY(Y=y) = 0 otherwise

or is this basically what you have? and why did you also add those extra functions in them?, I don't understand how your getting those functions from what i have...

 

[jbunniii]

ok i see, but i looked on wikipedia and for the bounoulli dist. and geometric dist. i thought it was

PX(X=1) = θ
PX(X=0) = 1-θ
PX(X=x) = 0 otherwise
PY(Y=y>=0) = θ(1-θ)^y
PY(Y=y) = 0 otherwise

or is this basically what you have? and why did you also add those extra functions in them?, I don't understand how your getting those functions from what i have...

Yes, your functions and mine are equivalent. I added the extra functions so I don't have to express them as individual cases (x = 1, x = 0, otherwise) as you did.

To see that mine are the same as yours, just plug in various values of $k$.

For example, if

$$b(k) = (1 - \theta)\delta(k) + \theta\delta(k-1)$$

then notice that $\delta(k)$ is zero except when $k = 0$, and $\delta(k-1)$ is zero except when $k = 1$.

Thus $b(0) = (1 - \theta)(1) + 0 = (1 - \theta)$ and $b(1) = 0 + \theta(1) = \theta$ and $b(k) = 0$ if $k$ is neither 0 nor 1.

Similarly with the geometric distribution.

The point is that it makes it possible to write a one-line expression that is valid for all $k$, which in turn makes it easier to express the convolution sum.

By the way, this isn't some weird invention of mine - it's a standard thing to do when working with functions defined in pieces, and the notation ($\delta(k)$ and $u(k)$) are quite standard as well.

 

[sneaky666]

i have one last concern about why my answer is wrong:

since the main equation is
$$\sum$$ P(X=k)P(Y=z-k) for all k
inf on top
k=-inf on bot

so how I got my answers is because of
PX(X=1) = θ
PX(X=0) = 1-θ
PX(X=x) = 0 otherwise
PY(Y=y>=0) = θ(1-θ)^y
PY(Y=y) = 0 otherwise

i have 3 cases, if X=k is 0, 1, or something else
so if k = 0 then you would have
P(X=k)P(Y=z-k)
P(X=0)P(Y=z-0)
(1-θ)P(Y=z)
(1-θ)θ(1-θ)^z
θ(1-θ)^(z+1)

so if k = 1 then you would have
P(X=k)P(Y=z-k)
P(X=1)P(Y=z-1)
θθ(1-θ)^(z-1)
θ^2(1-θ)^(z-1)

so if k != 0,1 then you would have
P(X=k)P(Y=z-k)
0*P(Y=z-k)
0

(and of course the summation beside them, i didnt add it here)

So i don't understand what is wrong here?

 

[jbunniii]

You are mixing up your $k$ and $z$. One of them is a dummy variable used in the summation, and the other one is the letter that you use to fill in the blank:

P(X + Y = ____)

So let's pick which one is which and stick with it.

If you want to fill in the blank with z,

$$P(X + Y = z) = \sum_{k=-\infty}^{\infty} P(X = k) P(Y = z - k)$$

then $k$ is the dummy variable in the sum (it doesn't appear on the left side at all). So your three cases apply to $z$, not $k$:

Case 1: z < 0
Case 2: z = 0
Case 3: z > 0

Try that and see if it helps.