# Thread: Expectation Problem

1. ## Expectation Problem

1. Show that if a and b are constants with P(a<X<b)=1, then a< E(X)<b.

2. Aces. A standard deck of 52 cards is shuffled and dealt. Let X1 be the number of cards appearing before the first ace, X2 the number of cards between the first and second ace (not counting either ace), X3 the number between the second and third ace, X4 the number between the third and forth ace, and X5 the number after the last ace. it can be shown that each of these random variables Xi had the same distribution, i=1,2,...,5, and you can assume this to be true.
a)Write down a formula for P(Xi=k), 0<k<48.
b)Show that E(Xi)=9.6. [hint: Do not use your answer to a).]
c)Are X1,...,X5 pairwise independent? Prove your answer.

(For all <, it is less than or equal than a number.)

2. Originally Posted by Yan
1. Show that if a and b are constants with P(a<X<b)=1, then a< E(X)<b.

For the upper limit:

$\displaystyle E(X)=\int_a^b x p(x) \ dx \le \int_a^b b p(x) \ dx=b\int_a^b p(x) \ dx=b$

since $\displaystyle P(a<X<b)=\int_a^bp(x)\ dx=1$ is given.

The same argument works for the lower limit.

CB

3. Originally Posted by CaptainBlack
For the upper limit:

$\displaystyle E(X)=\int_a^b x p(x) \ dx \le \int_a^b b p(x) \ dx=b\int_a^b p(x) \ dx=b$

since $\displaystyle P(a<X<b)=\int_a^bp(x)\ dx=1$ is given.

The same argument works for the lower limit.

CB
how about question #2?