0
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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).

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ I have added another line to my answer. @mXdX $\endgroup$ – Kavi Rama Murthy Jan 23 '19 at 5:35
  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.