Var[aX] = ![E[a^2X^2] - {E[aX]}^2](http://l.wordpress.com/latex.php?latex=E%5Ba%5E2X%5E2%5D+-+%7BE%5BaX%5D%7D%5E2&bg=ffffff&fg=000000&s=0 "E[a^2X^2] - {E[aX]}^2")
![a^2E[X^2] - a^2E[X]^2](http://l.wordpress.com/latex.php?latex=a%5E2E%5BX%5E2%5D+-+a%5E2E%5BX%5D%5E2&bg=ffffff&fg=000000&s=0 "a^2E[X^2] - a^2E[X]^2")
![a^2(E[X^2] - E[X]^2)](http://l.wordpress.com/latex.php?latex=a%5E2%28E%5BX%5E2%5D+-+E%5BX%5D%5E2%29&bg=ffffff&fg=000000&s=0 "a^2(E[X^2] - E[X]^2)")
![a^2 Var[X]](http://l.wordpress.com/latex.php?latex=a%5E2+Var%5BX%5D&bg=ffffff&fg=000000&s=0 "a^2 Var[X]")
March 26, 2007
Exercises C.3-9
Bernoulli distribution
Let E[X] = p.
Var[X] = 
![E[X^2]](http://l.wordpress.com/latex.php?latex=E%5BX%5E2%5D&bg=ffffff&fg=000000&s=0 "E[X^2]")
– ![E[X]^2](http://l.wordpress.com/latex.php?latex=E%5BX%5D%5E2&bg=ffffff&fg=000000&s=0 "E[X]^2")
, which is positive. Since Exercises C.3-6
Let I be the indicator random variable, which equal to 1 when X ≥ t.
tI ≤ |X|, so E[tI] ≤ E[X], but E[tI] = tE[I] = aPr(X ≥ t). Hence Pr(X ≥ t) ≤ E[ X ] / t
+ 
.Exercises C.3-3
Expected gain: (-1 * 125 + 1 * 3 * 25 + 2 * 3 * 5 + 3)/216 = -17/216
Exercises C.3-2
Expectation of the index of the maximum element: 
Expectation of the index of the minimum element:
Exercises C.3-1
Expectation of two dice’s sum = 7
Expectation of the maximum value = (6 * 11 + 5 * 9 +4 * 7 + 3 * 5 + 2 * 3 + 1 * 1) / 36 = 161/36
