# Poisson and Geometric Random Variable questions

#### dizzle1518

##### New Member
Hi all,

Can anyone tell me if I got the correct answers for the below two questions? Any help is appreciated.

Q:Let X have a Poisson distribution with parameter λ. If we know that P(X = 1|X ≤ 1) = 0.8, then what is the expectation and variance of X?

A:So, (X=0|X≤ 1)=0.2, which equals e^-λ=0.2. Taking the log of both sides and multiplying by -1 we get λ =1.60944=E(X)=VAR(X). Is this right?

Q:A random variable X is a sum of three independent geometric random variables with mean 4. Find the mean, variance, and moment generating function of X/4.

A: Each of the three geometric r.v.'s has E(X)=(1-p)/p=4. Solving for p we get 0.2. Since these are three independent geometric r.v.'s then their sum is a negative binomial r.v. with parameters r=3 and p=0.2. E(X) of a negative binomial r.v. is (r*(1-p))/p, which gives us 12 and VAR(X) (r*(1-p))/p^2=60. Are these correct?
I am not sure how to get the moment generating function of X/4. The moment generating function is defined as E(e^(tx)). In my notes I have that E(e^(atx)) gives us the moment generating function Mx(at). The moment generating function for neg. binomial r.v. is (p/(1-(1-p)e^t))^r. Since the constant a=1/4 would the moment generating function just be (p/(1-(1-p)e^t/4))^r?

Thanks,
--David

#### Dason

Q:Let X have a Poisson distribution with parameter λ. If we know that P(X = 1|X ≤ 1) = 0.8, then what is the expectation and variance of X?

A:So, (X=0|X≤ 1)=0.2, which equals e^-λ=0.2. Taking the log of both sides and multiplying by -1 we get λ =1.60944=E(X)=VAR(X). Is this right?
Where are you using the fact that you're conditioning on X $$\leq$$ 1?
Hint: It's important and what you have right now isn't correct. But it's a good start.

#### dizzle1518

##### New Member
Where are you using the fact that you're conditioning on X $$\leq$$ 1?
Hint: It's important and what you have right now isn't correct. But it's a good start.
Dason,

I am not sure what you mean by where am I using the fact that I am conditioning on X ≤ 1. Can you clarify?

Thanks

#### Dason

I mean you aren't using the fact that it's telling you that you know that X <= 1.

#### dizzle1518

##### New Member
I mean you aren't using the fact that it's telling you that you know that X <= 1.
I thought I was. Since the Poisson distribution is discrete over positive values of x and we are given that we know x <= 1 the only other possibility is that x = 0. The way I looked at it is, these are two mutually exclusive evens ie if x= 1 it cannot equal 0 and vice versa. Hence the .2 probability of x=0 conditioned on the fact that x<=1

#### dizzle1518

##### New Member
ok i got λ equals 4 by using the following relationship P(A|B) = P(A ∩ B)/P(B) and setting it up in the following way:

P(X = 1 ∩ X ≤ 1)=λe^-λ
P(X≤1)=e^-λ+λe^-λ

.8=λe^-λ/(e^-λ+λe^-λ)

is this right?

#### Dason

.2 = $$e^{-\lambda}/(e^{-\lambda} + \lambda e^{-\lambda}) \rightarrow .2 = 1/(1+\lambda)$$