Question

Suppose a random variable X has the following probability density function:f(x)=1/x, 1≤x≤C, or f(x)=0 otherwise a) what must the value of C be so that f(x) is a probability density function?b) find P(X<2)c) find E(X) and Var(X)

Answer 1

a)
`C=e^1=e`

b)
`P(X<2)=\int_(1)^2 (1)/(x) dx = ln(x) \Big|_1^2 =ln(2)-ln(1)=ln(2)=0.693`

c)
`E(X) =\int_(1)^e x (1)/(x) dx =x \Big|_1^e \ =e-1`

`E(X^2) =\int_(1)^e x^2 (1)/(x) dx =(x^2)/(2) \Big|_1^e \ =(1)/(2)(e^2 -1)`

`Var(X)=E(X^2)-[E(X)]^2= (1)/(2)(e^2 -1) -(e-1)^2 = 0.242`

Explanation:

a) what must the value of C be so that f(x) is a probability density function?

In order to be a probability function we need this condition:

`\int_(1)^C (1)/(x) dx =1`

And solving the left part of the integral we have:

`ln(x) \Big|_1^C \ =1`

`ln(C)-ln(1)=1`, so then
`C=e^1=e`

b) find P(X<2)

We can find this probability on this way using the density function:

`P(X<2)=\int_(1)^2 (1)/(x) dx = ln(x) \Big|_1^2 =ln(2)-ln(1)=ln(2)=0.693`

c) find E(X) and Var(X)

We can find the expected value on this way:

`E(X) =\int_(1)^e x (1)/(x) dx =x \Big|_1^e \ =e-1`

In order to find the Var(X) we need to find the second moment given by:

`E(X^2) =\int_(1)^e x^2 (1)/(x) dx =(x^2)/(2) \Big|_1^e \ =(1)/(2)(e^2 -1)`

And now we can use the following definition:

`Var(X)=E(X^2)-[E(X)]^2= (1)/(2)(e^2 -1) -(e-1)^2 = 0.242`