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.