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#1 2007-10-15 10:44:39

George74
Guest

Probability help

Hi all.  I was wondering if my work for the following was correct:

Suppose (X,Y) is a random vector with a probability mass function p_(X,Y) of (i, j) = 1/10 for 1 ≤ i ≤ j ≤ 4.

Compute the conditional expectation of Y given X.

So for this, I realized that the possibilities of (i,j) were (1,1), (1,2), (1,3), (1,4), (2,2), (2,3), (2,4), (3,3), (3,4), and (4,4). 

P(1,1) = .1
P(1,2) = .1
P(1,3) = .1
P(1,4) = .1
P(2,1) = 0
P(2,2) = .1
P(2,3) = .1
P(2,4) = .1
P(3,1) = 0
P(3,2) = 0
P(3,3) = .1
P(3,4) = .1
P(4,1) = 0
P(4,2) = 0
P(4,3) = 0
P(4,4) = .1

P(X=1) = .1 + .1 + .1 + .1 = .4
P(X=2) = .1 + .1 + .1 = .3
P(X=3) = .1 + .1 = .2
P(X=4) = .1

I did p(x,y)/P(X=x) for each pair of coordinates listed above that had a probability not equal to 0.   I got:
For (1,1), I got 1/4
For (1,2), I got 1/4
For (1,3), I got 1/4
For (1,4), I got 1/4
For (2,2), I got 1/3
For (2,3), I got 1/3
For (2,4), I got 1/3
For (3,3), I got 1/2
For (3,4), I got 1/2
For (4,4), I got 1.

Multiplying each of these probabilities by the value of the x coordinate, I got (1/4 * 1) + (1/4 * 1) + (1/4 * 1) + (1/4 * 1) + (1/3 * 2) + (1/3 * 2) + (1/3 * 2) + (1/2 * 3) + (1/2 * 3) + (1 * 4) = 10

So I got 10 for a conditional expectation of Y given X.

#2 2007-10-15 15:05:03

George74
Guest

Re: Probability help

Never mind!  I figured out that my method was wrong...

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