# Expected values of IID random variables

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)

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

$$E([X-\mu]^{2})$$, 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).
• Thanks. Having trouble understanding why $E[X^{2}] = (\sigma^{2} + \mu^{2})$, though. – mXdX Jan 23 '19 at 5:33
• @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. – Kavi Rama Murthy Jan 23 '19 at 7:23