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Currently stuck on this, I know I should probably use the mean deviation of the binomial distribution but I can't figure it out.

Glen_b
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thyde
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    Hi, welcome to CV. While questions such as this are welcome, we do treat them differently - if you put more information into your question, you can get hints and guidance. Please see the relevant paragraph in [his page of the help](http://stats.stackexchange.com/help/on-topic), and the guidelines at the `self-study` [tag wiki](http://stats.stackexchange.com/tags/self-study/info). Please add the `self-study` tag and modify your question as suggested (that is, show what you've tried, or at least explain what you know about expectations and binomials) and identify where your difficulties lie. – Glen_b Dec 14 '14 at 21:50
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    you might also look at jensen's inequality – seanv507 Dec 14 '14 at 22:47
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    @seanv507 certainly, if we use Jensen's inequality, that does it in one step, and if thyde has covered it that's all that would be needed, but in this instance there's a really elementary proof that is well within reach of students who only know some very basic properties of expectation and variance. – Glen_b Dec 14 '14 at 22:48
  • $E[Y^2] = Var[Y] + E[Y]^2$ which becomes $Var[X] + (E[X] - np)^2$, then solving we get: $npq + (np - np)^2 = npq$. Is this correct? – thyde Dec 14 '14 at 22:52
  • I have learned Jensen's inequality, but only in the context of convexity. Could you possibly demonstrate it's applicability here? – thyde Dec 14 '14 at 22:55
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    I think you are confusing yourself with Var. just use E. you need to show that $E|X-np|\le \sqrt{E[|X-np|^2]}$. – seanv507 Dec 14 '14 at 23:09

1 Answers1

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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).

  1. 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.

  2. 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.]

--

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.

Glen_b
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