What is the Expected Value of Discrete Random Variables?

[forty]

The distribution function of a random variable X is given by:

F(x) =

0 if x <-3
3/8 if -3 <= x < 0
1/2 if 0 <= x < 3
3/4 if 3 <= x <4
1 if x => 4

Calculate E(X) and E(X² - 2|X|)

Well I'm at a loss of E(X) although once I know this the other should be fairly simple..

Ive got E(X) = -3(3/8)-2(3/8)-1(3/8)+1(1/8)+2(1/8)+1(1/4)+2(1/4)+3(1/4)+4(1/4) = 5/8

Is this the right method?

Rather than making another post I also need a hand with the following.
Let X be the number of different birthdays among four persons selected randomly. Find E(X).

I know how to do this in principal E(X) = 0.p(0) + 1.p(1) + 2.p(2) + 3.p(3) + 4.p(4)

but i don't know how to find the probability mass function p(x) I know it will takes values 0,1,2,3,4 and will probably involve the pick formula 365!/(365-x)! or something along those lines.

Any help would be greatly appreciated!

 

[Billy Bob]

The distribution function of a random variable X is given by:

F(x) =

0 if x <-3
3/8 if -3 <= x < 0
1/2 if 0 <= x < 3
3/4 if 3 <= x <4
1 if x => 4

Calculate E(X) and E(X² - 2|X|)

Well I'm at a loss of E(X) although once I know this the other should be fairly simple..

Ive got E(X) = -3(3/8)-2(3/8)-1(3/8)+1(1/8)+2(1/8)+1(1/4)+2(1/4)+3(1/4)+4(1/4) = 5/8

Is this the right method?

No. You were given the cdf F(x), so first you could find the pmf p(x). It looks like you tried to do that, but not correctly. The only x-values that give a positive value for p(x) are x=-3, 0, 3, and 4.

Rather than making another post I also need a hand with the following.
Let X be the number of different birthdays among four persons selected randomly. Find E(X).

I know how to do this in principal E(X) = 0.p(0) + 1.p(1) + 2.p(2) + 3.p(3) + 4.p(4)

but i don't know how to find the probability mass function p(x) I know it will takes values 0,1,2,3,4 and will probably involve the pick formula 365!/(365-x)! or something along those lines.

Any help would be greatly appreciated!

You could treat finding p(x) as five different problems, for each of x=0, 1, 2, 3, and 4.

Example: p(1)=prob they all have same birthday=P(1st person has any b-day)*P(2nd is same)*P(3rd is same)*P(4th is same)=1*(1/365)*(1/365)*(1/365).

p(4)=prob they all have different b-days, which is similar.

For p(2), there may be clever ways, but a boring safe way is to make a tree of possibilities. Similarly for p(3).

If you use this approach (i.e. one person at a time), just remember the trick that for the first person, you always have P(1st person has any b-day)=1.

 

[forty]

p(x) = 3/8 for x = -3
p(x) = 1/8 for x = 0
p(x) = 2/8 for x = 3
p(x) = 2/8 for x = 4

is this right?

E(X) = -9/8 + 6/8 + 8/8 = 5/8 (this gives the same answer as above I presume this is just coincidence if my first method is wrong)

 

[Billy Bob]

p(x) = 3/8 for x = -3
p(x) = 1/8 for x = 0
p(x) = 2/8 for x = 3
p(x) = 2/8 for x = 4

is this right?

E(X) = -9/8 + 6/8 + 8/8 = 5/8 (this gives the same answer as above I presume this is just coincidence if my first method is wrong)

New method is right, and first was just coincidence.