1. ## sample mean question

Y_1,Y_2, . . . ,Y_n are independent and identically distributed Exp(Q)
random variables.
Which are the distribution of the sample mean?

Sample mean=X=(1/n)*SIGMA(Y_k)

1. X~Exp(Q)
2. X~Exp(nQ)
3. X~Exp(Q/n)
4. X~Gamma(n,Q)
5. X~Gamma(n,(Q/n))

2. Originally Posted by MC Squidge
Y_1,Y_2, . . . ,Y_n are independent and identically distributed Exp(Q)
random variables.
Which are the distribution of the sample mean?

Sample mean=X=(1/n)*SIGMA(Y_k)

1. X~Exp(Q)
2. X~Exp(nQ)
3. X~Exp(Q/n)
4. X~Gamma(n,Q)
5. X~Gamma(n,(Q/n))

So, what attempts have you made?

3. the sum is distributed as $\Gamma(n,Q)$ by using MGFs
Next make a calc one change of variable to divide that rv by n.

4. Im sorry can you do one of them as an example I dont quite understand.

5. let $W=Y_1+\cdots +Y_n$

Then by using MGFs $W\sim\Gamma(n,Q)$

So, $X=W/n$ and $f_X(x)=f_W(w)\bigl|{dw\over dx}\bigr|$

SO, start by writing the density of W, which there are two ways, that's one reason I didn't
write my gamma density.

6. Ok thanks a lot for your help I think Im beggining to understand it now. So 1,2,3 and 5 AREN'T distributions of the sample mean because their MGFs are different, but 4 is because its MGF is the same?

edit; actually its 5 that has the same MGF as the sample mean right? not 4

edit 2; yes ive definitely figured it out its 5 only. tell me if I`ve got it. thanks a lot for the help