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 dens…

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asked Jul 24, 2020 172k views

1 Answer

Augusto Hack · Answer 1 · 2020-07-29T01:14:26+0000

Answer:

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