Can You Give Me Help On Expected Value Please?

Question

The Question: Let $X_{1}, X_{2}, ..., X_{9} $ be a random sample of size 9 from a normal distribution $N(2,4)$. Let $Y_{1}, Y_{2} , Y_{3}, Y_{4}$ be an independent random sample from a normal distribution $N(1,1)$. Let $\bar{X}$ and $\bar{Y}$ be their sample means respectively. Let's assume that $\bar{X}$ and $\bar{Y}$ be independent. Find $E(\bar{X} - \bar{Y})$ and $Var(\bar{X} - \bar{Y})$.

My Attempt: So for $E(\bar{X} + \bar{Y})$,

$E(\bar{X} - \bar{Y})$= $E(\bar{X}) - E(\bar{Y})$ = $E(\sum_{i=1}^{9} \frac{X_{i}}{9})$ - $E(\sum_{j=1}^{4} \frac{Y_{j}}{4})$ = $\frac{1}{9} \sum_{i=1}^{9} E(X_{i}) - \frac{1}{4} \sum_{j=1}^{4} E(Y_{j})$ = $\frac{1}{9} (2*9) - \frac{1}{4} (4*1) $ = $1$.

For $Var(\bar{X} + \bar{Y})$

Since $\bar{X}$ and $\bar{Y}$ are independent, we can write the above statement as $Var(\bar{X} - \bar{Y}) = Var(\bar{X}) - Var(\bar{Y})$. By definition, $Var(\bar{X}) - Var(\bar{Y})= Var(\sum_{i=1}^{9} \frac{Y_i}{9}) - Var(\sum_{j=1}^{4} \frac{Y_{j}}{4}) = \frac{4}{9} - \frac{1}{4} = \frac{7}{36}$.

I hope I did this correctly. Am I on the right track? Just give me some hints please. Also, how would I know if $\bar{X} - \bar{Y}$ is normal?

Answer 1

Your calculation for the mean value is correct. As for the variance, you have that

$$\text{var}(\bar{X}-\bar{Y}) = \text{var}(\bar{X})\color{red}+\text{var}(\bar{Y}).$$

Notice the plus on the right hand side.

You have in fact that $\bar{X}$ has the distribution $N(2,\frac{4}{9})$, and that $\bar{Y}$ has the distribution $N(1,\frac{1}{4})$. Since they are independent, their difference will have the distribution $N(2-1,\frac{4}{9}+\frac{1}{4}) = N(1,\frac{4}{9}+\frac{1}{4})$.

Answer 2

1) The expectation \begin{align*} E[\bar X - \bar Y] &= E[\bar X]-E[\bar Y]\\ &=\frac{1}{9}E[X_1+\dotsb+X_9]-\frac{1}{4}E[Y_1+\dotsb+Y_4]\\ &=\frac{1}{9}(9)E[X_1]-\frac{1}{4}(4)E[Y_1]\\ &=2-1\\ &=1. \end{align*}

2) The variance \begin{align*} \text{Var}[\bar X-\bar Y]&=\text{Var}[\bar X]+\text{Var}[\bar Y]\\ &=\text{Var}\left[\frac{X_1+\dotsb+X_9}{9}\right]+\text{Var}\left[\frac{Y_1+\dotsb+Y_4}{4}\right]\\ &=\frac{1}{81}(9)\text{Var}[X_1]+\frac{1}{16}(4)\text{Var}[Y_1]\\ &=\frac{1}{9}(4)+\frac{1}{4}(1)\\ &=\frac{25}{36}. \end{align*} In general, if $X\sim N(\mu, \sigma^2)$, then $$\bar X\sim N(\mu, \sigma^2/n).$$ Thus, if you sample $n$ from $X$, and $m$ from $Y\sim N(\nu, \tau^2)$, then $$\bar X+ \bar Y\sim N(\mu+\nu, \sigma^2/n+\tau^2/m)$$ and $$\bar X -\bar Y\sim N(\mu-\nu,\sigma^2/n+\tau^2/m).$$