Question

Let X be given by its distribution function F(x) , such that
f(x) = 0 if x<=0
f(x) = 1/16 x^4, if 0 < x <=2
f(x) = 1 , if x > 2

1. Find E(X), Var(X), SD(X)
2. Graph the density function
3. Graph the distribution function F(x)

Answer 1

\[f(x)=\frac{dF(x)}{dx}\]\[E(X)=\int\limits_{-\infty}^{\infty}xf(x)dx\]\[Var(X)=E(X^2)-[E(X)]^2=\int\limits_{-\infty}^{\infty}x^2f(x)dx-[E(X)]^2\]\[SD(X)=\sqrt{Var(X)}\]

Answer 2

Because \(\int\limits_{a}^{b}f(x)dx=F(b)-F(a)\) it's easier to do these integrations using the distribution function.

Answer 3

i am confused about infinity. would E(X) use 1/16x^4 or would it be 1? and would the variable be 2 and 0 instead of infinity?

Answer 4

The infinity is only there because it's possible for a distribution to have a value of f(x) for every value of x (an example is the normal distribution). Because in your case if x<0 or x>2, the value of f(x) is zero, the value of the integral outside these limits is zero, so you don't need to calculate it.

Answer 5

\[\int\limits_{\infty}^{\infty}x(1/16x)^2dx\]

Answer 6

would this be E(X)

Answer 7

No because they've given you its distribution function and not it's density function. First you need to differentiate the distribution function to find the density function:\[f(x)=\frac{dF(x)}{dx}=\frac{d}{dx}(\frac{1}{16}x^4)=\frac{1}{4}x^3\]\[E(X)=\int\limits_{-\infty}^{\infty}xf(x)dx=\int\limits_{0}^{2}\frac{1}{4}x^4dx\]

Answer 8

ok i understand now. you would need the distribution to obtain the density. so in order to get VAR(X) i would take (1/4x^4)^2?

Answer 9

which would be 1/4x^5?

Answer 10

Umm, not quite. So far we have:\[E(X)=\int\limits_{0}^{2}\frac{1}{4}x^4dx=[x^5/20]^2_{0}=\frac{32}{20}=\frac{8}{5}\]

Answer 11

So then you'd square that and take it away from \(E(X^{2})\).

Answer 12

ok im just getting myself more confused. the other problem was different but it was for density function, so this requires me to obtain the density function and them calculate the distribution

Answer 13

Yes, to calculate E(X), Var(X), SD(X), you need to use the density function which you obtain by differentiating the distribution function.

Answer 14

ok i understand now. i have alot of work to do

Answer 15

Haha, good luck :)

Answer 16

thanks