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$X_1,.., X_n$ are independent, identical distributed random variables. They are continuous, too. Let $$\bar{X}= \frac{1}{n} \cdot (X_1+..+X_n)$$

Determine the expected value and variance of $\bar{X}$ if $$X_i \text{ is normally distributed, } X_i \sim N(\mu, \sigma^2)$$

I don't know how do it good?

I check expected value of normal distribution on internet. This is

$$E(X) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty}xe^{\frac{-x^2}{2}}dx$$

Now I try form it so I have good solutin

$$-\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{\frac{-x^2}{2}}d\left(-\frac{x^2}{2}\right)= -\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}= 0$$

It work like this? But no idea about variance..?

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1 Answer 1

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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}

As a bonus, $X \sim N(\mu, \sigma^2/n)$.

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  • $\begingroup$ Thank you very much!! I do all wrong because I just handed it in.. But better I have solution and understand for exam because can ask similar question. Can pls also show for this? math.stackexchange.com/questions/2547210/… I don't understanded answer of other people $\endgroup$
    – eyesima
    Commented Dec 4, 2017 at 0:10

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