# E(X)===> Help with Expectation problem

#### snthorne

##### New Member
Assume the length X in minutes of a particular type of telephone conversation is random variable with probability density function

f(x)={1/5 e^-x/5 x>0
{0 otherwise

Determine the mean length=E(X) of this type of telephone conversation.

Here is the work I did:
Integration by parts where: u=x dv=e^-x/5 dx
du=dx v=-5e^-x/5

E(X)= -∞ ∫ ∞ xf(x) <-- where integrated from -∞ to ∞
=1/5 0 ∫ ∞ xe^-x/5 dx<-- integrated from 0 to ∞
=-1/5 xe^-x/5 - ∫-5e^-x/5 dx
=-1/5 xe^-x/5 + 5 ∫ e^-x/5 dx
=-1/5 xe^-x/5 + ∫ e^u du
=-1/5 xe^-x/5 - e^-x/5 0|∞

#### vinayan

##### New Member
To find the expctation of the given density function, you need to integrate the function
x * (1/5) * e^(-x/5) from 0 to infinity (as you have shown), which is a gamma integral, after
1/5 is taken outside of the intgral, with m=2 and p =(1/5).

The form of gamma integral is,
integral from 0 to infinity e^(-mx) * x^(p-1) = Gamma(p) / m^p
where Gamma(p) =(p-1)!.
Note that n!=1*2*3*...n

So here the answer is (1/5) * Gamma(2) / (1/5)^2 = 5.
The first 1/5 shown above is the one which is outside the integral sign.

You can solve using the formula for integration of product of two functions also.

The product formula for integration is integral of two functions will be first fn. * derivative of the second fn. minus integral of (derivative of the first fn. * integral of the second).

#### jerryb

##### New Member
snthorne said:
=-1/5 xe^-x/5 + 5 ∫ e^-x/5 dx
=-1/5 xe^-x/5 + ∫ e^u du
=-1/5 xe^-x/5 - e^-x/5 0|∞
in your second line here u= (-x/5) right?

so du = -1/5dx let [ be the needed intergral, i don't know how you got the integral sybol

5[e^(-x/5)dx

5*(-5)[e^(-x/5)(-1/5)dx

-25[e^u du

and finish up
jerry