mean and variance...is this correct???

Hi,

Suppose that X1 and X2 are independent discrete random variable and each assume the values 0,1,and 2 with equal probability. Find the probability of X=1/2(X1+x2) What is the mean of its probability distribution

this is what I have so far...I am not sure if this is correct:

X1 0 1 2 X2= 0,1,2
P(x1) 1/3 1/3 1/3 P(x2) 1/3 1/3 1/3

p(x1+X2)= 1/9, 1/9, 1/9
1/2(X1+x2)= 1/18, 1/18, 1/18

E(x)= 0(1/18)+1(1/18)+2(1/18)

Also, let X1 and X2 be independent standard normal variable find the mean and variance...I am not sure how to do this

Thanks

No good. The entire distribution of X1 + X2 consists of nine elements:

0 + 0 = 0
0 + 1 = 1
0 + 2 = 2
1 + 0 = 1
1 + 1 = 2
1 + 2 = 3
2 + 0 = 2
2 + 1 = 3
2 + 2 = 4

You seem to be missing some.

so probability when X1 =1 and X2=1 is P(X1+X2)= 1/3 * 1/3 ? and then do same thing for all nine ?

Also, when it comes to mean and variance, whats the difference between discrete and normal distribution?

thank you so much

Notice that the sum of your probabilities must be unity. In your original version, you have 1/9 only 3 times. That won't do. Nine times is better.