$$
\begin{bmatrix}*&*&*\\*&*&*\\*&*&*\end{bmatrix}
$$

$var(\overline{x})=var(\frac{1}{3n}\sum_{i=1}^{3}\sum_{j=1}^{n}X_{ij})$
$=\frac{1}{9n^{2}}\sum_{i=1}^{3}var(\sum_{j=1}^{n}X_{ij})$
$=\frac{1}{3n^{2}}E[\sum_{j=1}^{n}(X_{ij}-\mu )]^{2}$
$=\frac{1}{3n^{2}}\sum_{j=1}^{n}\sum_{j'=1}^{n}E[(X_{ij}-\mu)(X_{ij'}-\mu)]$
$=\frac{1}{3n^{2}}\sum_{j=1}^{n}E[(X_{ij}-\mu)^{2}]+2\sum_{j=1}^{n}\sum_{j'=j+1}^{n}E[(X_{ij}-\mu)(X_{ij'}-\mu)]$

$$
\begin{bmatrix}*&*&*\\ 0&0&*\\ *&*&*\end{bmatrix}
$$

$$
\begin{bmatrix}4&*&*\\ 0&0&0\\ *&*&*\end{bmatrix}
$$
I win

$$
\begin{bmatrix}*&*&*\\ *&1&*\\ *&*&*\end{bmatrix}
$$
by not putting $0$

$$
\begin{bmatrix}1&*&*\\ *&1&0\\ *&0&1\end{bmatrix}
$$
let's make $I_3$

$$
\begin{bmatrix}1&*&*\\ 2&1&0\\ 4&0&1\end{bmatrix}
$$

$$
\begin{bmatrix}1&4&0\\ 2&1&0\\ 4&0&1\end{bmatrix}
$$

$$
\begin{bmatrix}*&*&*\\*&*&*\\*&*&*\end{bmatrix}
$$

$$
\begin{bmatrix}*&*&*\\*&*&1\\*&*&1\end{bmatrix}
$$

$$
\begin{bmatrix}*&*&*\\*&2&1\\*&2&1\end{bmatrix}
$$

$$
\begin{bmatrix}3&*&1\\*&2&1\\*&2&1\end{bmatrix}
$$

$$
\begin{bmatrix}3&5&1\\*&2&1\\3&2&1\end{bmatrix}
$$

(Took me ages to understand because my bias of the yau is 又)
有神快拜 (yau5 sun4 faai3 baai3)

Well, I guess I missed the opportunity to ask Balarka about my question then

Hi guys. My teacher gave me te following exercise: Show
$$
\mathbb P(X+Y\geq x+y)\leq\mathbb P(X\geq x)+\mathbb P(Y\geq y).
$$
I don't remember if she specified anything about $X$ and $Y$; I'm guessing all we know is they are random variables. If I knew whether they were discrete of continous, I would probably use the convolution formula. I'm guessing I'll just have to use the non-decreasing property of $\mathbb P$ here, as I don't see any better option. Could someone give me a hint to begin?