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Bivariate discrete random variable

Yankel

Active member
Jan 27, 2012
398
Hello

I am trying to solve this problem:

A coin is given with probability 1/3 for head (H) and 2/3 for tail (T).
The coin is being drawn N times, where N is a Poisson random variable with E(N)=1. The drawing of the coin and N are independent. Let X be the number of heads (H) in the N draws. What is the correlation coefficient of X and N ?

So I started this by creating a table as if it was a finite problem, just to see how it behaves, but it didn't lead me too far. Since there is independence, every event P(X=x , N=n) is equal to P(X=x|N=n)*P(N=n). So this is like a tree diagram sample space. In order to find the correlation, I need the covariance and the variances. The variance of N, it's easy, 1. How do I find the rest of the stuff ?

Thanks !
 

zzephod

Well-known member
Feb 3, 2013
134
Hello

I am trying to solve this problem:

A coin is given with probability 1/3 for head (H) and 2/3 for tail (T).
The coin is being drawn N times, where N is a Poisson random variable with E(N)=1. The drawing of the coin and N are independent. Let X be the number of heads (H) in the N draws. What is the correlation coefficient of X and N ?

So I started this by creating a table as if it was a finite problem, just to see how it behaves, but it didn't lead me too far. Since there is independence, every event P(X=x , N=n) is equal to P(X=x|N=n)*P(N=n). So this is like a tree diagram sample space. In order to find the correlation, I need the covariance and the variances. The variance of N, it's easy, 1. How do I find the rest of the stuff ?

Thanks !
You have \(\bar{N}\), \(\sigma_N\) and the joint distribution, so:

\(\displaystyle \bar{X} = \sum_{n=0..\infty, x=0,..n} x f_{X,N}(x,n)=\sum_{n=0..\infty} \frac{n}{3}f_N(n)=\frac{1}{3}\bar{N}\)

\(\displaystyle \sigma^2_X= \sum_{n=0..\infty, x=0,..n} (x-\bar{X})^2 f_{X,N}(x,n)=\sum_{n=0..\infty}\frac{2n}{3}f_N(n)=\frac{2}{3}\bar{N}\)

\(\displaystyle {\rm{Cov}}(X,N)= \sum_{n=0..\infty, x=0,..n} (x-\bar{X})(n-\bar{N}) f_{X,N}(x,n)=\sum_{n=0..\infty}\frac{(n-\bar{N})^2}{3}f_N(n)=\frac{\sigma^2_N}{3}\)

so:

\(\displaystyle \rho_{X,N}=\frac{{\rm{Cov}}(X,N)}{\sigma_X \sigma_N}=\ ...\)

The key idea here is that for the double summation you can always choose to do that over \(x\) first.

.
 
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