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

[math25]

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

 

[tkhunny]

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.

 

[math25]

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

 

[tkhunny]

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.