I'm trying to make sure I am understanding the following properties:

For $X_1, \ldots, X_n$ IID random variables with mean $\mu$ and standard deviation $\sigma$,

$E(c) = c$, where $c$ is any constant

$E(X_i) = \mu$,

$E(\mu) = \mu$ (from the first equality)

$E(X_i - \mu) = E(X_i)-E(\mu) = \mu-\mu = 0$

Are these correct? Additionally, I'd like to think about

$E([X-\mu]^{2})$, and

($\sum_{i=1}^{k} [X-\mu])^{3}$

Where $X$ is any of the $X_i$'s (since they are identically distributed, I can say $X_i = X$, for all $i$, correct?)

Self-studying these concepts, any guidance very much appreciated.


What you have done is correct. $E(X-\mu)^{2}=E(X^{2}-2\mu X+\mu^{2})=EX^{2} -2\mu EX+\mu^{2} =(\sigma^{2}+\mu^{2})-2\mu^{2} +\mu^{2}=\sigma^{2}$. In general $E(X-\mu)^{2}$ is always the variance of $X$. T0 calculate $E( \sum\limits_{i=1}^{k}(X_i-\mu))^{3}$ you have to expand $(\sum\limits_{i=1}^{k}X_i-\mu)^{3}$ as $\sum\limits_{i=1}^{k}\sum\limits_{j=1}^{k}\sum\limits_{l=1}^{k} (X_i -\mu)(X_j-\mu)(X_l-\mu)$. [ I have used the following: $\sigma^{2}=EX^{2}-\mu^{2}$ by definition . Hence $EX^{2}=\mu^{2}+\sigma^{2}$]. How do we compute $E(X_i -\mu)(X_j-\mu)(X_l-\mu)$?. The value is $0$ if any two of $i,j,l$ are distinct and it is $E(X-\mu)^{3}$ if $i=j=l$. You can calculate $E(X-\mu)^{3}$ by expanding the cube (using Binomial Theorem).

  • $\begingroup$ Thanks. Having trouble understanding why $E[X^{2}] = (\sigma^{2} + \mu^{2})$, though. $\endgroup$ – mXdX Jan 23 '19 at 5:33
  • $\begingroup$ Also, that last part is the cube of the entire sum, not just the summand. $\endgroup$ – mXdX Jan 23 '19 at 5:34
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    $\begingroup$ I have added another line to my answer. @mXdX $\endgroup$ – Kavi Rama Murthy Jan 23 '19 at 5:35
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    $\begingroup$ @mXdX Sorry that I ddin't read the last part correctly. I have revised my answer. $\endgroup$ – Kavi Rama Murthy Jan 23 '19 at 5:38
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    $\begingroup$ @mXdX There are $k$ terms in the sum with $i=j=k$. So you will get $kE(X-\mu)^{3}$. For the fourth power you can use a similar argument. $\endgroup$ – Kavi Rama Murthy Jan 23 '19 at 7:23

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