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The following easy proof is well known to all students of statistics. I studied it three years ago and at present I can not remember two steps of it. If you can help me a little.

$$E({X_i}^2) = \mu^2 + \sigma^2$$ $$E(\bar{X}^2) = \mu^2 + \frac{\sigma^2}{n}$$

How to prove the above two results. You may give me some hints or the complete proof. I prefer the last one.

The main result with it proofs is mentioned below.

Thank you for your answer.

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Supriyo
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1 Answers1

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Hint: You just need the following facts:

  1. $Var(cX)=c^2Var(X)$ for some constant $c$
  2. $Var(X)+(EX)^2=E(X^2)$
  3. $Var(\sum X_i)=\sum Var(X_i)+\sum_i\sum_{j\neq i}Cov(X_i,X_j)$
  4. For independent random variables $X,Y$, $Cov(X,Y)=0$

Try using these facts. Otherwise,

Complete Proof: $Var(\bar{X})=Var\left(\frac{1}{n}\sum_{i=1}^nX_i\right)=\frac{1}{n^2}Var(\sum_{i=1}^nX_i)=\frac{1}{n^2}\sum_{i=1}^nVar(X_i)=\frac{\sigma^2}{n}$

QED
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