Find mean and variance of a random vector

[Dustinsfl]

Homework Statement

The components of a random vector $\mathbf{X} = [X_1, X_2, \ldots, X_N]^{\intercal}$ all have the same mean $E_X[X]$ and the same variance $var(X)$. The "sample mean" random variable
$$
\bar{X} = \frac{1}{N}\sum_{i = 1}^NX_i
$$
is formed. If the $X_i$'s are independent, find the mean and variance of $\hat{X}$. What happens to the variance as $N\to\infty$? Does this tell you anything about the PMF of $\bar{X}$ as $N\to\infty$?

Homework Equations

The Attempt at a Solution

Since the $X_i$'s are independent,
$$
\hat{X} = \frac{1}{N}\sum_{i = 1}^NE_{X_i}[X_i]
$$
but the \(X_i\)'s have the same means, thus, all are the same so
\begin{align*}
&= \frac{1}{N}NE[X]\\
&= E[X]
\end{align*}
Additionally, independence implies that $cov(X_i, X_j) = 0$ for $i\neq j$ so the variance is
\begin{align*}
var\Big(\sum var(X_i)\Big)
&= \sum_{i = 1}^Nvar(X_i)\\
&= \sum_{i = 1}^N\big(E[X_i^2] - E^2[X_i]\big)\\
&= \sum_{i = 1}^N\big(E[X_i^2]\big) - NE^2[X]
\end{align*}
Can the variance be simplified any more? I am not sure what happens to the variance as N tends to infinity either.

 

[Ray Vickson]

Homework Statement

The components of a random vector $\mathbf{X} = [X_1, X_2, \ldots, X_N]^{\intercal}$ all have the same mean $E_X[X]$ and the same variance $var(X)$. The "sample mean" random variable
$$
\bar{X} = \frac{1}{N}\sum_{i = 1}^NX_i
$$
is formed. If the $X_i$'s are independent, find the mean and variance of $\hat{X}$. What happens to the variance as $N\to\infty$? Does this tell you anything about the PMF of $\bar{X}$ as $N\to\infty$?

Homework Equations

The Attempt at a Solution

Since the $X_i$'s are independent,
$$
\hat{X} = \frac{1}{N}\sum_{i = 1}^NE_{X_i}[X_i]
$$
but the \(X_i\)'s have the same means, thus, all are the same so
\begin{align*}
&= \frac{1}{N}NE[X]\\
&= E[X]
\end{align*}
Additionally, independence implies that $cov(X_i, X_j) = 0$ for $i\neq j$ so the variance is
\begin{align*}
var\Big(\sum var(X_i)\Big)
&= \sum_{i = 1}^Nvar(X_i)\\
&= \sum_{i = 1}^N\big(E[X_i^2] - E^2[X_i]\big)\\
&= \sum_{i = 1}^N\big(E[X_i^2]\big) - NE^2[X]
\end{align*}
Can the variance be simplified any more? I am not sure what happens to the variance as N tends to infinity either.

You are complicating things unnecessarily. If $\mu = E\, X_i$ and $\sigma^2 = \text{Var}\, X_i$ $\forall i$, then
$$E\, \sum_{i=1}^n X_i = n \mu, \; \text{Var} \, \sum_{i=1}^n X_i = n \sigma^2\\<br /> \text{so}\\<br /> E\, \bar{X} = \mu, \; \text{Var}\, \bar{X} = \frac{\sigma^2}{n}.$$
Leaving the variance in this last form is the usual way these things are done, and yields the most insight into the problems's structure.

 

[Dustinsfl]

You are complicating things unnecessarily. If $\mu = E\, X_i$ and $\sigma^2 = \text{Var}\, X_i$ $\forall i$, then
$$E\, \sum_{i=1}^n X_i = n \mu, \; \text{Var} \, \sum_{i=1}^n X_i = n \sigma^2\\<br /> \text{so}\\<br /> E\, \bar{X} = \mu, \; \text{Var}\, \bar{X} = \frac{\sigma^2}{n}.$$
Leaving the variance in this last form is the usual way these things are done, and yields the most insight into the problems's structure.

Ok so that means the variance goes to zero then but what does that mean for the PMF?