In: Statistics and Probability
Let A be a non-negative random variable (A>0)
a) If A is discrete, show that E[A] >= 0
b) If A is continous , show E[X]>=0
a)
A discrete random variable A has a countable number of possible values.
Example: Two dice are thrown. Let A represent the sum of two dice. In this case, A is non negtative.
Then the probability distribution of X is as follows:
A |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
11 |
12 |
P (A) |
1/36 |
2/36 |
3/36 |
4/36 |
5/36 |
6/36 |
5/36 |
4/36 |
3/36 |
2/36 |
1/36 |
E(A) = ∑ A∙ P(A)
= [ 2∙(1/36) ] + [ 3∙(2/36) ] + [ 4∙(3/36) ] + [ 5∙(4/36) ] + [ 6∙(5/36) ] + [ 7∙(6/36) ] + [ 8∙(5/36) ] +
[ 9∙(4/36) ] + [ 10∙(3/36) ] + [ 11∙(2/36) ] + [ 12∙(1/36) ]
= [ 2 + 6 + 12 + 20 + 30 + 42 + 40 + 36 + 30 + 22 + 12 ] / 36
= 252 / 36
E(A) = 7
b)
Let A = Continuous random variable and it is defined as
A = x2 ; 0 < x < 3
= 0 ; Otherwise
E[x] =
=
=
= 1/4 { ( x4 )3 - ( x4 )0 }
= 1 / 4 [ 34 ]
E[X] = 81 / 4
E[X] = 20.25
In both the cases Expected value is greater than zero.