Poisson probability distribution conditional

sillybuffalo

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a) Let X have a Poisson probability distribution with mean λ=6. Once you observe X, toss a fair coin X times and let Y be the number of heads that show. Find the probability that Y=4.

b) Let the Poisson random variable X above have an arbitrary mean λ>0 and find the probability that Y=y for every integer y≥0, where once again Y is the number of heads in X throws of a fair coin.


I'm really lost with this problem, but to start with Part A, I think that I'm supposed to do something along the lines of...
P(Y=4|X) * P (X = some value)

I believe that Y has a binomial probability distribution (since we're flipping coins) and X has a Poisson probability distribution (as stated in the problem). Any further insight?
 
a) Let X have a Poisson probability distribution with mean λ=6. Once you observe X, toss a fair coin X times and let Y be the number of heads that show. Find the probability that Y=4.

b) Let the Poisson random variable X above have an arbitrary mean λ>0 and find the probability that Y=y for every integer y≥0, where once again Y is the number of heads in X throws of a fair coin.


I'm really lost with this problem, but to start with Part A, I think that I'm supposed to do something along the lines of...
P(Y=4|X) * P (X = some value)

I believe that Y has a binomial probability distribution (since we're flipping coins) and X has a Poisson probability distribution (as stated in the problem). Any further insight?
What you have suggested looks good - so time to get to work. For part (a) you will calculate for a specific λ (6) and a specific Y (4). You need to organize your work so that part (b) is systematic. Are you allowed perhaps to use a spreadsheet?

P(Y=4 | X) is a binomial with n=X, p=q=0.5, m=Y=4. For X > Y,
P(Y=4 | X) = (0.5)^X C(X, Y)
Note that this distribution is independent of λ.

P(X) is independent of Y. X is an integer and λ>0 is real.
P(X) = λ^X e^(-λ) / X!
 
What you have suggested looks good - so time to get to work. For part (a) you will calculate for a specific λ (6) and a specific Y (4). You need to organize your work so that part (b) is systematic. Are you allowed perhaps to use a spreadsheet?

P(Y=4 | X) is a binomial with n=X, p=q=0.5, m=Y=4. For X > Y,
P(Y=4 | X) = (0.5)^X C(X, Y)
Note that this distribution is independent of λ.

P(X) is independent of Y. X is an integer and λ>0 is real.
P(X) = λ^X e^(-λ) / X!

I'm having trouble simplifying and finding a value for Part A.

This is the summation I have written down:
anotherproblem4_1.gif
Does this look set up correctly?

 
Last edited:
I'm having trouble simplifying and finding a value for Part A.

This is the summation I have written down:
Does this look set up correctly?

Yes. [BTW, the LaTeX code for infinity is "\infty".]

The two numbers raised to power x might be combined:

\(\displaystyle \displaystyle P(Y\!=\!4) = \sum_{x=4}^{\infty} \binom x 4 \dfrac{e^{-6}\ 3^x}{x!} \)

The way the problem is written makes me think they want numerical values. In that case I would make a spreadsheet with x in one column and the corresponding term in the next column. Keep increasing x (adding rows) till the size on the next term is negligible.

For part B, does it look like you can replace 4-->Y and 6-->\(\displaystyle \lambda\) in your formula?
 
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