# Math Help - help with discrete random variables

1. ## help with discrete random variables

1. The problem statement, all variables and given/known data

1.
Suppose u flip a coin
Z = 1 if the coin is heads
Z = 3 if the coin is tails
W = Z^2 + Z
a)
what is the probability function of Z?
b)
what is the probability function of W?

2.
Let Z ~ Geometric (theta). Compute P(5<=Z<=9).

2. Relevant equations

3. The attempt at a solution

1.
what i think
a)
3 if tail
0 otherwise
b)
12 if tail
0 otherwise

I don't see how the probability part comes into this, where do i include the 1/2 chance part...

or maybe its:
a)
Pz(Z) = 1/2 if Z is 1 or 3
0 otherwise
b)
Pw(W) = 1/2 if W is 2 or 12
0 otherwise

2.

2. Z can be equal to 1 or 3.
The probability distribution function of Z will tell you what is the probability that Z takes each of these 2 values.

and P(Z=2)=P(outcome is tail)=1/2

Similar reasoning applies to W.

3. I dont understand this question.

Let Z ~ Geometric(theta). Compute P(5<=Z<=9).

How do i do this?

4. X ~ $\mbox{Geometric}(p)$ has the probability Distribution:

$P(X=x) = p^{x-1} (1 - p)^{x}$

You should now be able to compute P(5<=Z<=9)!!

5. I am still very confused..

I dont understand what values p should be and what values x should be...

6. p is the parameter of the distribution (same as the theta in your question).
X is the variable we look at (Z in your question)
So P ( 5<=Z<=9) = P(Z=5)+P(Z=6)+P(Z=7)+P(Z=8)+P(Z=9)

Z ~ $Geo(\theta)$

$P(Z=z)= {(1-\theta)}^{(z-1)} * \theta$

$P(Z=5)= {(1-\theta)}^{(5-1)} * \theta$

$P(Z=6)= {(1-\theta)}^{(6-1)} * \theta$

$P(Z=7)= {(1-\theta)}^{(7-1)} * \theta$

$P(Z=8)= {(1-\theta)}^{(8-1)} * \theta$

$P(Z=9)= {(1-\theta)}^{(9-1)} * \theta$

7. from your previous post, isn't it supposed to be

P(Z=z)=( theta^(z-1) ) * (1 - theta)^z
where z is 5,6,7,8,9

or is this exactly what you have but simplified?

8. Originally Posted by Sneaky
from your previous post, isn't it supposed to be

P(Z=z)=( theta^(z-1) ) * (1 - theta)^z
where z is 5,6,7,8,9

or is this exactly what you have but simplified?
It is:
$P(Z=z)={(1-\theta)}^{(z-1)}*\theta$
exactly what I wrote in my post.

Check your notes for the way geometric distribution is defined: Z ~ Geo ( $\theta$), then P(Z=z) = probability of obtaining a success after z-1 failures.