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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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    @Yuval Filmus: Someone has! But the original poster wrote explicitly he had no trouble with that part, and seemed to be only concerned with the variance of the difference between sample mean and population mean.2011-04-23

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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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    @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.2011-04-23