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

Answer:

a)  $C=e^1=e$

b) $P(X$

c) $E(X) =\int_{1}^e x \frac{1}{x} dx =x \Big|_1^e \ =e-1$

$E(X^2) =\int_{1}^e x^2 \frac{1}{x} dx =\frac{x^2}{2} \Big|_1^e \ =\frac{1}{2}(e^2 -1)$

$Var(X)=E(X^2)-[E(X)]^2= \frac{1}{2}(e^2 -1) -(e-1)^2 = 0.242$

Step-by-step 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:

$\int_{1}^C \frac{1}{x} dx =1$

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:

$P(X$

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

We can find the expected value on this way:

$E(X) =\int_{1}^e x \frac{1}{x} dx =x \Big|_1^e \ =e-1$

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

$E(X^2) =\int_{1}^e x^2 \frac{1}{x} dx =\frac{x^2}{2} \Big|_1^e \ =\frac{1}{2}(e^2 -1)$

And now we can use the following definition:

$Var(X)=E(X^2)-[E(X)]^2= \frac{1}{2}(e^2 -1) -(e-1)^2 = 0.242$