[ Average values ] $[$ Classical inequalities (other).]

There are $n$ random vectors of the form ( $y_{1}, y_{2}, y_{3}$ ), where exactly one random coordinate is equal to 1, and the others are

0. They are added together. The resulting random vector is $\boldsymbol{a}$ with coordinates $\left(Y_{1}, Y_{2}, Y_{3}\right)$.

a) Find the expected value of the random variable $\boldsymbol{a}^{2}$.

b) Prove that $|\boldsymbol{a}| \geq \frac{n}{\sqrt{3}}$.
a) The quantity $Y_{j}$ - the number of units among the numbers $y_{j}$ - is distributed according to the binomial law with probability $p=1 / 3$ and the number of trials . Therefore,
$\mathrm{E} Y_{j}=n / 3$ and $\mathrm{D} Y_{j}=n \cdot 1 / 3 \cdot 2 / 3=2 n / 9$.

Thus, $\mathrm{E}\left(\boldsymbol{a}^{2}\right)=\mathrm{E}\left(Y_{1}^{2}+Y_{2}^{2}+Y_{3}^{2}\right)=3\left(\mathrm{D} Y_{1}+\mathrm{E}^{2} Y_{1}\right)=\frac{2 n+n^{2}}{3}$.

b) $\sqrt{Y_{1}^{2}+Y_{2}^{2}+Y_{3}^{2}}=\sqrt{\frac{Y_{1}^{2}+Y_{2}^{2}+Y_{3}^{2}}{3}} \cdot \sqrt{3} \geq \frac{Y_{1}+Y_{2}+Y_{3}}{3} \cdot \sqrt{3}=\frac{n}{\sqrt{3}}$.

## Answer

a) $\frac{2 n+n^{2}}{3}$.