David..

Please help me in soving this with details.
  y=1   y=2   y=3
x=1   0.05   0.05   0.10
x=2   0.05   0.10   0.15
x=3   0.15   0.15   0.20

Use the above joint probability distribution to answer the foll:
1. The expected value of Y is closest to

a. .2
b. 1
c. 2.3
d. 2.2

2. Suppose Y is equal to 2, the probability that X is equal to 1 is closest to

a. .05
b. .25
c. .17
d. .2

3. The variance of X is closest to

a. .54
b. .61
c. .67
d. .74

Thanks

venkat,

I input into XLS here:
http://sheet.zoho.com/public/btzoho/0919-prob

This question is *great* for giving practice to conditional vs. unconditional ... I would recommend “using” this problem to meditate on conditional/unconditional/joint…note the variance is E(X^2) - (E(X))^2, which comes in handy often!

Thanks, David

I suppose that Q2 is about P(X=1|Y=2) so the answer is a) => .05. Right?

Reg
Sebastian

Hi Sebastian, not quite (that’s why this is a handy exercise)

joint probability P(X=1,Y=2) = 0.05
unconditional P(Y = 2) = 0.3
conditional or marginal P(X=1|Y=2) = =.05/.30 = 0.17

David

OK, i get it.
It’s X=1 when Y=2, not X=1 and Y=2.

Thx
Sebastian