Distribution Function and Properties of Continuously Distributed Variances

[kasse]

X is continuously distributed with probability density

f_{X}(x) = nx^{n-1}, if 0 < x \leq 1
and
f_{X}(x) = 0, otherwise

Find the distribution function F(x) of X. Find the probability that X lies between 0.25 and 0.75 when n=1 and when n=2. Find the median of X, i.e. the value of a so that P(X \leq a) = 1/2, when n=1 and when n=2. Find E(X) when n=1 and when n=2 and compare with the corresponding medians.

I'm first going to try to find the distribution function. Does this simply mean finding n in the probability density?

 

[statdad]

If f(x) is the density function for a random variable, the distribution function is

<br /> F(x) = \int_{-\infty}^x f(t) \, dt<br />

You can calculate P(a \le X \le b) either by

<br /> F(b) - F(a)<br />

or

<br /> \int_a^b f(x) \, dx<br />

(these are actually the same things dressed up in different notations)

To find the median solve F(a) = 0.5

To find the expected values calculate integrate x f(x).

 

[kasse]

<br /> F(x) = \int_{-\infty}^x nt^{n-1} \, dt = [t^n]^{x}_{- \infty}<br />

Shoult the integral limits be from 0 to x instead of negative infinity to x? I don't know, but that's the way it's supposed to be done if I interppret an example in my book correctly. Then I get

<br /> F(x) = x^n<br />

and

P(0.25 < X < 0.75) = 0.5 for both n=1 and n=2.

and the medians a=0.5 and a=0.707 when n=1 and n=2, respectively

and finally

E(X) = 0.5 and E(X) = 2/3 when n=1 and n=2, respectively.

 

[statdad]

Yes - I gave the general definitions for your needs. Since your density is zero outside of the interval [0,1], every

<br /> \int_{-\infty}^\infty \text{ stuff} \, dx<br />

reduces to

<br /> \int_0^1 \text{stuff} \, dx<br />