# Conditional expected value

1. May 19, 2013

### ParisSpart

we have a variable that takes values 1,...,6 with density:
n 1 2 3 4 5 6
f(n) 0.1 0.2 0.1 0.3 0,176 0,124

What is the average price (expected value) of X under the condition that X is even?

E(X/X=even)=k*P(X=k/X=even)=0.2*2+4*0.3+0.124*6
i am doing this but its says its not correct what i am doing wrong in the type?

2. May 19, 2013

### I like Serena

Use the product rule:

P(X=k|X even) = P(X=k) P(X even)

3. May 19, 2013

### ParisSpart

P(X even) here is 1/2? or 1/3?

4. May 19, 2013

### I like Serena

P(X even) = P(X=2 or X=4 or X=6)

Use the sum rule for mutually exclusive events:

P(A or B) = P(A) + P(B) if A and B are mutually exclusive.

5. May 19, 2013

### ParisSpart

P(X=2)=0.2? from the table?

6. May 19, 2013

### I like Serena

Yes...

7. May 19, 2013

### ParisSpart

but its says that its not correct my answer... i find this 1.462656 in the final answer

8. May 19, 2013

### I like Serena

What did you calculate then?

9. May 19, 2013

### ParisSpart

first i found this P(X=2 OR X=4 OR X=6)=0.624
AND E(X/X=even)=k*P(X=k/X=even)=0.2*2*0.624+0.624*4*0.3+0.124*6*0.624...=1.4626

10. May 19, 2013

### I like Serena

Oops. My mistake.
The rule is: P(X=k|X even) = P(X=k) / P(X even)

11. May 19, 2013

### ParisSpart

k*(P(X=k)/P(X even)) like this?

12. May 19, 2013

Yes.