Probability theory - Poisson and Geometric Random Variable questions

[dizzle1518]

Homework Statement [/b]

There are two problems I need help with, which are posted below. Any help is appreciated.

1)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?

2)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.


The attempt at a solution[/b]

1)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?


2)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

 

[vela]

Homework Statement [/b]

There are two problems I need help with, which are posted below. Any help is appreciated.

1)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?

2)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.


The attempt at a solution[/b]

1)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?

No. That would work if you had P(X=0), but you have P(X=0 | X≤1). You need to account for the conditional probability.

 

[dizzle1518]

No. That would work if you had P(X=0), but you have P(X=0 | X≤1). You need to account for the conditional probability.

Can you give me a hint on how to account for it? Do I use the law of total probability and and set this up along the following lines:

P(X=1) = P(X=1|X=0)*P(X=0)+P(X=1|X=1)*P(X=1)
= 0*P(X=0)+1*P(X=1)

The right hand is 0.8, so we get P(X=1)=0.8. And then just solve for lambda?

 

[vela]

Use the definition of conditional probability, P(A|B) = P(A ∩ B)/P(B). I found λ=4.

 

[dizzle1518]

Use the definition of conditional probability, P(A|B) = P(A ∩ B)/P(B). I found λ=4 .

How did you get λ=4? P(X = 1|X ≤ 1) = 0.8=P(X = 1 ∩ X ≤ 1)/P(X ≤ 1). Since we are given that X is ≤ 1 then P(X ≤ 1)= 1 right? So that just gives us P(X = 1 ∩ X ≤ 1) = 0.8, which is basically saying P(X = 1) = 0.8 since x cannot be 1 and 0 at the same time. this sets up e^-λ=.8. But that doesn't equal 4

 

[vela]

No, P(A ∩ B) and P(B) aren't conditional probabilities.

 

[dizzle1518]

No, P(A ∩ B) and P(B) aren't conditional probabilities.

ok so P(B) in this case is P(X ≤ 1) right? so doesn't that equal 1 since we are given that X ≤ 1? then we get 1*P(X = 1|X ≤ 1)=P(X = 1 ∩ X ≤ 1) =1*.8=P(X = 1 ∩ X ≤ 1)
do i have this set up correct? what's the next step? I'm lost.

 

[vela]

ok so P(B) in this case is P(X ≤ 1) right? so doesn't that equal 1 since we are given that X ≤ 1?

No, P(X≤1) is not a conditional probability. You don't assume X≤1 in calculating it.

 

[dizzle1518]

No, P(X≤1) is not a conditional probability. You don't assume X≤1 in calculating it.

But I am not calculating P(X≤1). I am calculating P(X = 1 ∩ X ≤ 1). you said to use P(A|B) = P(A ∩ B)/P(B).

P(A|B) we have. that's P(X = 1|X ≤ 1) = 0.8
P(A ∩ B) = P(X = 1 ∩ X ≤ 1) = ?
P(B) = P(X≤1)=?

thus we have:
0.8=P(X = 1 ∩ X ≤ 1)/P(X≤1)...is this set up correct until now? if so what's the next step?

 

[vela]

But I am not calculating P(X≤1). I am calculating P(X = 1 ∩ X ≤ 1). you said to use P(A|B) = P(A ∩ B)/P(B).

P(A|B) we have. that's P(X = 1|X ≤ 1) = 0.8
P(A ∩ B) = P(X = 1 ∩ X ≤ 1) = ?
P(B) = P(X≤1)=?

thus we have:
0.8=P(X = 1 ∩ X ≤ 1)/P(X≤1)...is this set up correct until now? if so what's the next step?

Come up with expressions for P(X = 1 ∩ X ≤ 1) and P(X≤1) in terms of λ.

 

[dizzle1518]

Come up with expressions for P(X = 1 ∩ X ≤ 1) and P(X≤1) in terms of λ.

ok i got λ equals 4 by 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?

 

[vela]

Yes!

 

[dizzle1518]

Yes!

thanks for all your help