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Let $X_1,..,X_n$ be independent, identically distributed (continuous) random variables. Let $$\bar{X} = \frac{1}{n} \cdot \left(X_1+...+X_n\right)$$

Determine the expected value and variance of $\bar{X}$ if

(i) $X_i$ is equally distributed on $(0,1)$, $X_i \sim R(0,1)$

(ii) $X_i$ exponentially distributed, $X_i \sim E(\lambda)$

Hi maths people I think teacher can ask this kind of question in test next week. But not sure how solve it good?

(i) Because they say continuous, we can use formula for variance.

$$E(\bar{X}) = \frac{a+b}{2} = \frac{0+1}{2}= \frac{1}{2}$$

$$Var(\bar{X}) = E(\bar{X}^2) - (E(\bar{X}))^2 = \frac{1}{12}(b-a)^2 = \frac{1}{12}(1-0)^2 = \frac{1}{12}$$ I do it correct for this? How do it correct for (ii)? I have no idea because no find formula..

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  • $\begingroup$ Not quite. Remember if you add random variables with variance $v$, then it will also add. So for the sum of all the $n$ $X$'s you will have a variance of $nv$. But then when you divide by some number then the variance is divided by that number squared so it becomes $\frac{nv}{n^2}$ = $\frac{v}{n}$. $\endgroup$ – stuart stevenson Dec 2 '17 at 11:08
  • $\begingroup$ If you are saying that $X_i$ are equally distributed over the open interval between 0 and 1, and $X_i \sim R(0, 1)$ then, $X_i$ is undefined as it can never be 0 or lower and the expected value can never be 0. $\endgroup$ – Harsh Parekh Dec 4 '17 at 0:29
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Let $X_i \sim R(\mu,\sigma^2)$ and $X_i$'s are i.i.d.

And $X = (1/n)\sum_i X_i$.

Then by linearity of expectation, \begin{align} \mathbb{E}[X] &= \frac{1}{n} \sum_i \mathbb{E}[X_i]\\ &= \frac{1}{n} (n\mu) = \mu \end{align}

As $X_i$ are i.i.d., \begin{align} \text{Var}[X] &= \frac{1}{n^2}\sum_i \text{Var}[X_i]\\ &= \frac{1}{n^2}(n\sigma^2)\\ &= \sigma^2/n \end{align}

This holds for all i.i.d. $X_i$.

Now you can simply substitute in your values, also, I recommend that you try to understand the procedure and do the proof of things that are used yourself. For example try to prove why variance of independent variables can be simply added or why do we multiply with the square of the constant. These are simple proofs that come directly from the definitions.

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Not quite. Remember if you add random variables with variance $v$, then it will also add. So for the sum of all the $n$ $X$'s you will have a variance of $nv$. But then when you divide by some number then the variance is divided by that number squared so it becomes $\frac{nv}{n^2}$ = $\frac{v}{n}$.

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    $\begingroup$ "if you add random uncorrelated variables" $\endgroup$ – Gabriel Romon Dec 2 '17 at 11:24
  • $\begingroup$ I don't understand how will look it? $\endgroup$ – roblind Dec 2 '17 at 11:36

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