Given N-distribution, calculate expected value and Var of a function

Question

$X_{{i}}=X_{{1}}...X_{{n}}$ is an iid. random variable with the distribution $N \left( {\frac {\alpha}{\beta}},{\frac {{\alpha}^{4}}{{\beta}^{4}}} \right) $

I would like to calculate the expectation value as well as variance of the following function:

$Y={\frac {{\beta}^{2}\sum _{i=1}^{n}X_{{i}}}{n{\alpha}^{2}}}-{\frac { \beta}{na}}+1 $

I believe I can use the iid. quality of $X$ to get its expectation value:

$E \left( X \right) ={\frac {\sum _{i=1}^{n}X_{{i}}}{n}}={\frac {\alpha}{\beta}}$

Which when used in substitution results in the following expectation value of $Y$:

${\it E} \left( Y \right) ={\frac {\beta}{\alpha}}-{\frac {\beta} {n\alpha}}+1$

1. Is my result for $E(Y)$ correct?
2. How do I proceed from here and calculate $Var(Y)$?

Thanks.

EDIT: I apologize for the small LaTeX font. I'm not sure how to fix it.

Answer 1

Answer:

Using the linearity of expectations

$Y={\frac {{\beta}^{2}\sum _{i=1}^{n}X_{{i}}}{n{\alpha}^{2}}}-{\frac { \beta}{na}}+1 $ $E(Y) = E\left({\frac {{\beta}^{2}\sum _{i=1}^{n}X_{{i}}}{n{\alpha}^{2}}}-{\frac { \beta}{na}}+1\right)$

$E(Y) = \frac{n\cdot\beta^2\cdot\alpha}{n\cdot{\alpha}^2\cdot \beta}-\frac{\beta}{n\cdot\alpha} + 1$

If you simplify, you get the result and it is correct..

Do the same thing with Variance given the fact all the random variables are iid, $ Var(X_1) + \cdots + Var(X_n) = Var(Y)$ and the property $Var(cX) =c^2Var(X)$

Thanks

Satish

Answer 2

you have in Y the sum of a normal distribution and rest just constants. We know that the sum of iid $N(\mu,\sigma^2)$ variables is $N(n\mu,n\sigma^2)$ distributed (see http://en.wikipedia.org/wiki/Sum_of_normally_distributed_random_variables).

Therefore we get:

$E(Y)=\frac{\beta^2n\alpha}{\beta n\alpha^2}-\frac{\beta}{n\alpha}+1=\frac{\beta}{\alpha}-\frac{\beta}{n\alpha}+1$

For the variance we use $Var(c\sum X_i)=c^2n\frac{\alpha^4}{\beta^4}$ (with $c=\frac{\beta^2}{n\alpha^2}$) and Variance of the constants is $0$, to get:

$V(Y)=\frac{1}{n}$