# Show that $\mathbb P\left(\left | \frac{Y_1+\dots +Y_N}{N} - \bar{Y} \right | \geq \lambda \right) \leq \frac{\mathbb{V}[Y]}{{\lambda}^{2}N}$

I am working on showing that $$\mathbb P\left(\left | \frac{Y_1+\dots +Y_N}{N} - \bar{Y} \right | \geq \lambda \right) \leq \frac{\mathbb{V}[Y]}{{\lambda}^{2}N}.$$ I am given that $Y_1, Y_2, \dots, Y_N$ are independent, identically distributed random variables. I have already proved that $$\mathbb{P}(\left | X \right | \geq \lambda ) \leq \frac{\mathbb{V}[X]}{{\lambda}^{2}},$$ so essentially I just need to show that $$\frac{\mathbb{V}[\frac{Y_1+\dots +Y_N}{N} - \bar{Y}]}{{\lambda}^{2}} = \frac{\mathbb{V}[Y]}{{\lambda}^{2}N}.$$

I have expanded the numerator the following way:

\begin{align*}\mathbb{V}\left [ \frac{Y_1+\dots +Y_N}{N} - \bar{Y} \right ] &= \mathbb{E} \left [ \left ( \frac{Y_1+\dots +Y_N}{N} - \bar{Y} \right )^{2} \right ] - \left ( \mathbb{E} \left [ \frac{Y_1+\dots +Y_N}{N} - \bar{Y} \right ] \right )^{2}\\ &= \mathbb{E} \left [ \frac{(Y_1+\dots +Y_N)^{2}}{N^{2}} \right ] - 2\frac{\bar{Y}}{N}\mathbb{E}(Y_1+\dots +Y_N) + \bar{Y}^{2} \\ &- \left ( \frac{1}{N}\mathbb{E}(Y_1+\dots +Y_N) -\bar{Y} \right )^{2}\\ &= \frac{\mathbb{E}(Y_1+ \dots + Y_N)^{2}}{N^{2}} - \frac{2\bar{Y}}{N} \mathbb{E}(Y_1+ \dots + Y_N)+\bar{Y}^{2}\\ & - \frac{ \left [ \mathbb{E}(Y_1+\dots +Y_N) \right ]^{2}}{N^{2}} + \frac{2\bar{Y}}{N}\mathbb{E} (Y_1+\dots +Y_N) - \bar{Y}^{2}\\ &= \frac{\mathbb{E}(Y_1+ \dots + Y_N)^{2}}{N^{2}} - \frac{ \left [ \mathbb{E}(Y_1+\dots +Y_N) \right ]^{2}}{N^{2}} \end{align*} in the numerator. So now I get $$\frac{\frac{\mathbb{E}(Y_1+ \dots + Y_N)^{2}-\left [ \mathbb{E}(Y_1+\dots +Y_N) \right ]^{2}}{N^{2}}}{\lambda^{2}} = \frac{\mathbb{E}(Y_1+ \dots + Y_N)^{2}-\left [ \mathbb{E}(Y_1+\dots +Y_N) \right ]^{2}}{\lambda^{2}N^{2}}.$$ Even if the expression in the numerator is $\mathbb{V}[Y]$ (which I am not sure about), it is still being divided by $\lambda^{2}N^{2}$ and not by $\lambda^{2}N$ as it should in $\frac{\mathbb{V}[Y]}{N{\lambda}^{2}}$. What is it that I am doing wrong? Thanks.

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The expression you currently have in the numerator of the last displayed formula is the variance of $Y_1+Y_2+\cdots+Y_N$. For how to get from this to the right thing, please see my answer below. – André Nicolas Apr 23 '11 at 6:04
I used $n$ instead of $N$, because $n$ looks better. Instead of your "$Y$-bar" I used $\mu$. In the formula just below "I have expanded ...", the last term that you subtract is superfluous, it is $0$, but it is not a mistake. But the argument is incomplete, you have not at all shown the relationship between the expression in the last displayed formula and the variance of one (and hence all) of the $Y_i$. – André Nicolas Apr 23 '11 at 6:18
@user6312: Thanks. Could you hint me on how to get to that relationship from my expression? Is it even possible? – user9967 Apr 23 '11 at 6:25
@CrazyFormula: Expand $(Y_1+Y_2+\cdots+Y_n)^2$. This is sort of like expanding $(a+b)^2$. You get terms $Y_1^2+\cdots +Y_n^2$, and the sum of all "mixed" terms $2Y_iY_j$ where $i < j$. Use the fact that $E(Y_iY_j)=E(Y_i)E(Y_j)=\mu^2$. {This is where you use independence, the expectation of a product of two independent random variables is the product of the expectations.) Then there is a bit of algebra, which I am sure you can handle. – André Nicolas Apr 23 '11 at 6:42
@CrazyFormula: But it is more pleasant to look at $(Y_1+\cdots +Y_n)-n\mu$, rewrite as $(Y_1-\mu)+\cdots +(Y_n-\mu)$, and then expand, use the expectation of an independent product rule to conclude that the expectations of all "mixed" terms is $0$. The long calculation that you did is then not used at all. – André Nicolas Apr 23 '11 at 6:56

First we state some basic results that are fairly easy to establish, and do not involve the complicated computations you have performed.

(a) If $X$ is a random variable, and $k$ is a constant, then the variance of $kX$ is $k^2$ times the variance of $X$. Also, if $a$ is any constant, the variance of $X+a$ is equal to the variance of $X$.

(b) If $U$ and $V$ are independent, then the variance of $U+V$ is the sum of the variances of $U$ and $V$. This then easily extends by induction to longer sums.

So now let $Y_1, Y_2,\dots,Y_n$ be independent, each with variance $\sigma^2$ (they need not be identically distributed).

Let $W=Y_1+ Y_2+\cdots +Y_n$.

Then the variance of $W$ is the sum of the variances of the $Y_i$, so it is $n\sigma^2$.

Now by (a) the variance of $W/n$ is $(1/n^2)(n\sigma^2)$, which is $\sigma^2/n$. And by the second assertion in (a), if each $Y_i$ has mean $\mu$, the variance of $W/n -\mu$ is also $\sigma^2/n$.

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the assignment tells me to use $\frac{Y_1+\dots +Y_N}{N} - \bar{Y}$ in order to prove this, so I would appreciate if you could point to a mistake in my calculations if I made one, or to help me simplify the expressions in the end! But of course your points make sense. – user9967 Apr 23 '11 at 6:04
@CrazyFormula: For an answer to the above comment, please look just below your post. I have the careless habit of putting comments in the wrong place. But while I am at it, might as well add that when you were asked to use what you said you were, the instructor probably then wanted you to compute the variance by using "standard" facts like the ones I have quoted. These are probably in your book or notes before the assignment. – André Nicolas Apr 23 '11 at 6:30