# Thread: probability-discrete randon variables, pmfs

1. ## probability-discrete randon variables, pmfs

I've never even covered this stuff before guys :S The below numbers where in a table but I couldnt recreate it on here , hope you clever people can figure it out

Discrete random variables X and Y have joint probability mass function:

Y=0 & X=0 =>0.2
Y=1 & X=0 =>0.2
Y=2 & X=0 =>0.1
Y=0 & X=1 =>0.2
Y=1 & X=1 =>0.3
Y=2 & X=1 =>0.0

(a) Find the marginal means and variances of both X and Y .
(b) Find the conditional mean of Y given X = 1.
(c) Find the correlation between X and Y

2. To get the marginal distribution of just one rv, you need to sum over the other one.
For example P(Y=0)=P(Y=0 and X=0) + P(Y=0 and X=1)=.4
You are summing over all values of x
P(Y=1)=P(Y=1 and X=0) + P(Y=1 and X=1)=.5
Since these probabilities sum to one
we have .1 left over for P(Y=2)
The last probability, Y=3 & X=1 =>0 should be erased.
Anything that has probably zero shouldn't be listed.
From here you should be able to get Y's mean and variance.
As for the conditional mean of Y when X=1,
you need the distribution of Y when X=1.
You only need these two
Y=0 & X=1 =>0.2
Y=1 & X=1 =>0.3
P(Y=0|X=1)=P(Y=0and X=1)/P(X=1)=.2/.5=.4
Thus P(Y=1|X=1)=1-P(Y=0|X=1)=.6
So, E(Y|X=1)=(0)(.4)+(1)(.6)=.6
The correlation between two rvs is the covariance divided by the st deviations
AND by the Cauchy-Schwarz inequality it has to be between -1 and 1.