Mean and variance of the sum This problem showed up on my exam:
A box contains 10 slips bearing the numbers $1,2,...,10$. We take out $2$ slips at random without replacement and add the two corresponding numbers. Find the mean and the variance of the sum.
I tried to solve it directly (using the definition of mean and variance) on the exam, but its fairly long, and I am trying to think of a smarter and faster idea to solve this. Does anyone have a better solution?
 A: A direct method:
Let $X_1,X_2,...,X_{10}$ be the result of all $10$ draws. It is an exchangeable sequence so
$Var(X_1+...+X_{10})=10\sigma^2+2{10 \choose 2}\sigma^2\rho$
where $\sigma^2$ is the variance of any $X_k$ and $\rho$ is the correlation between any pair of the X's.
$X_1+...+X_{10}$ is a constant, $55,$ so $Var(X_1+...+X_{10})=0$ from which we can obtain $\rho=-1/9.$  
Then $E(X_1^2)=(1/10)\sum_{i=1}^{10}i^2=\frac{11(21)}{6}$ and $\sigma^2=E(X_1^2)-5.5^2=33/4$ which gives us
$Var(X_1+X_2)=2\sigma^2+2\sigma^2\rho=44/3.$
A: Let the first number chosen be $X$ and the second be $Y$, and let $S = X+Y$ be their sum.  Then $$\mathrm{E}[S] = \mathrm{E}[\mathrm{E}[S \mid X]] = \sum_{x=1}^{10} \mathrm{E}[S \mid X = x]\Pr[X = x] = \frac{1}{10}\sum_{x=1}^{10} (x+\mathrm{E}[Y \mid X = x]).$$  To find this last expectation, we note that if $X = x$, then the expected value of $Y$ is simply the average of the remaining numbers not equal to $x$; i.e., $$\mathrm{E}[Y \mid X = x] = \frac{1}{9}\left(\frac{10(11)}{2} - x\right).$$  The rest is straightforward.
To find the variance, we write $$\mathrm{Var}[S] = \mathrm{E}[\mathrm{Var}[S \mid X]] + \mathrm{Var}[\mathrm{E}[S \mid X]].$$  To calculate the conditional variance, we can apply the formula $$\mathrm{Var}[Y \mid X] = \mathrm{E}[Y^2 \mid X] - \mathrm{E}[Y \mid X]^2.$$  The rest is similar to the expectation calculation above.
A: For the mean: define $X_c = 11 - X$, $Y_c = 11 - Y$.
Then $X + Y + X_c + Y_c = 22,$ and therefore 
$E(X + Y) + E(X_c + Y_c) = E(X + Y + X_c + Y_c) = 22.$
But by symmetry, the joint distribution of $X_c$ and $Y_c$ is identical to the joint distribution of $X$ and $Y;$ in particular, $E(X + Y) = E(X_c + Y_c).$
Therefore $2E(X + Y) = 22$ and $E(X + Y) = 11.$
A: For the mean, note you can make five pairs each adding to $11$, so the mean of one is $5.5$ and the.mean of two is $11$. For the variance, maybe you know $\sum_{i=1}^n i^2=n(n+1)(2n+1)/6$ but otherwise I don't have a quick way.
A: $E[X+Y] = E[X] + E[Y] \\ Var[X+Y] = Var[X]+Var[Y] + 2 Cov[X,Y]$
$E[X] = \frac 1{10} \sum_{x=1}^{10} x = \frac{1}{10} \frac{10\cdot 11}{2} = \frac {11} 2$
$E[X^2] = \frac 1{10} \sum_{x=1}^{10} x^2 = \frac{1}{10} \frac{10\cdot 11\cdot 21}{6} = \frac {77} 2$
$E[XY] = \frac{1}{10\cdot 9}(\underbrace{\sum_{x=1}^{10} (x \sum_{y=1}^{10} y - x^2)}_{\text{sum when $x\neq y$}}) = \frac 1{90} \sum_{x=1}^{10}(55x-x^2) = \frac{2640}{90} = 88/3$
$E[Y]=E[X], E[Y^2]=E[X^2]$, by symmetry
Thus:
$$E[X+Y] = 11$$
$$Var[X+Y] = (E[X^2]-E[X]^2)\times 2 + 2\times(E[XY]-E[X]^2) \\ = 2\times(\frac{77}{2}-\frac{121}{2}+\frac{88}{3})
\\ = \frac{44}{3}$$
