Finding mean of Y if Y=(X1+X2+X3)/3 given mean and variance of x's

[ellyezr]

Let X1, X2, X3 be three random variables. Suppose all three have mean μ and variance 1. The sample mean is Y = (X1 +X2 +X3)/3.
(a) Can you compute the mean of Y? If so, what is it? If not, why not?

I have that it is either μ OR that it is not possible to find, since we don't know if they are independent or not (as it says later in the question). I have a strong feeling that it is the latter, but I am not sure.

(b) If we assume that the three random variables are independent, what would the variance of Y be?
1/3 right? Just to be sure.

 

[Simon Bridge]

You should have notes on how to combine distributions.
How did you calculate the values you have suggested?

 

[ellyezr]

y=(x1+x2+x3)/3
x1 = x2 = x3 = mu
y=(mu+mu+mu)/3
y=3mu/3=mu

or can you not do that because you don't know if they are independent or not?

and no, I don't notes on that - trust me, I looked before I posted.

 

[Simon Bridge]

OK - well the first part seems to be saying you know nothing about the distributions on the ground that they are random. However, since the means are the same, does it make a difference?

The second part says they are independent - but nothing else - does that matter?

Or is the context of the problem important for figuring out what it all means?

Since you have no notes on this, you should try looking some up.
It would help me help you if I knew what level this should be answered at and if this forms part of a formal course.

 

[mathman]

Let X1, X2, X3 be three random variables. Suppose all three have mean μ and variance 1. The sample mean is Y = (X1 +X2 +X3)/3.
(a) Can you compute the mean of Y? If so, what is it? If not, why not?

I have that it is either μ OR that it is not possible to find, since we don't know if they are independent or not (as it says later in the question). I have a strong feeling that it is the latter, but I am not sure.

(b) If we assume that the three random variables are independent, what would the variance of Y be?
1/3 right? Just to be sure.

a) Averaging is linear - dpendence is irrelevant. E(Y) = μ
b) Yes.

 

[ellyezr]

thank you mathman