Show that if $X \sim Bin(n, p)$, then $E|X - np| \le \sqrt{npq}.$ Currently stuck on this, I know I should probably use the mean deviation of the binomial distribution but I can't figure it out.
 A: So that the comment thread doesn't explode I'm collecting my hints toward a completely elementary proof (you can do it shorter than this but hopefully this makes each step intuitive). I've deleted most of my comments (which unfortunately leaves the comments looking a little disjointed).


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*Let $Y=X-np$. Note $E(Y)=0$. Show $\text{Var}(Y)=npq$. If you already know $\text{Var}(X)$, you could just state $\text{Var}(Y)$, since shifting by a constant does nothing to variance.

*Let $Z=|Y|$. Write an obvious inequality in $\text{Var}(Z)$, expand $\text{Var}(Z)$ and use the previous result.  [You may want to slightly reorganize this into a clear proof, but I am attempting to motivate how to arrive at a proof, not just the final proof.]
That's all there is to it. It's 3 or 4 simple lines, using nothing more complicated than basic properties of variance and expectation (the only way the binomial comes into it at all is in giving the specific form of $E(X)$ and $\text{Var}(X)$ - you could prove the general case that the mean deviation is always $\leq \sigma$ just as readily).
[Alternatively, if you're familiar with Jensen's inequality, you can do it slightly more briefly.]
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Now that some time has passed, I'll outline a little more detail about how to approach it:
Let $Z=|X-nq|$. Then $\text{Var}(Z)=E(Z^2)-E(Z)^2$, and $E(Z^2)=E[(X-nq)^2]$ ...
Note that variances must be positive. The result follows.
