N Nas Apr 2010 11 0 May 24, 2010 #1 Does anyone have a clue on how to do this? A random variable x with a probability density function f(x):=| 1/(b-a) | 0 a≤x≥b x<a or x>b is said to be uniforly distributed over the interval (a,b) a) Find the mean of x b) Find the variance of x

Does anyone have a clue on how to do this? A random variable x with a probability density function f(x):=| 1/(b-a) | 0 a≤x≥b x<a or x>b is said to be uniforly distributed over the interval (a,b) a) Find the mean of x b) Find the variance of x

Random Variable May 2009 959 362 May 24, 2010 #2 EDIT: \(\displaystyle \mu = \frac{1}{2} \int^{b}_{a} x \ dx = \frac{1}{b-a} \frac{x^{2}}{2}|^{b}_{a} = \frac{1}{b-a} \frac{b^{2}-a^{2}}{2} = \frac{b+a}{2}\) Last edited: May 24, 2010

EDIT: \(\displaystyle \mu = \frac{1}{2} \int^{b}_{a} x \ dx = \frac{1}{b-a} \frac{x^{2}}{2}|^{b}_{a} = \frac{1}{b-a} \frac{b^{2}-a^{2}}{2} = \frac{b+a}{2}\)

N Nas Apr 2010 11 0 May 24, 2010 #3 thank you for the help. In order to calculate the variance, do I use the formula where the variance is equal to the integral of (x-mean)^2*f(x)dx?

thank you for the help. In order to calculate the variance, do I use the formula where the variance is equal to the integral of (x-mean)^2*f(x)dx?

Random Variable May 2009 959 362 May 24, 2010 #4 EDIT: \(\displaystyle \sigma^{2} = \frac{1}{b-a} \int^{b}_{a} (x-\mu)^{2} \ dx = \frac{1}{b-a} \int^{b}_{a} \Big(x-\frac{b+a}{2}\Big)^{2} \ dx\) \(\displaystyle = \frac{1}{b-a} \int^{b}_{a} \Big( x^{2} - (b+a) x + \frac{(b+a)^{2}}{4} \Big) \ dx \) \(\displaystyle = \frac{1} {b-a} \Big( \frac{x^{3}}{3} - (b+a) \ \frac{x^{2}}{2} + \frac{(b+a)^{2}}{4} \ x \Big)\Big|^{b}_{a} \) \(\displaystyle = \frac{1}{b-a} \Big(\frac{b^{3}}{3} - (b+a) \frac{b^{2}}{2} + \frac{(b+a)^{2}}{4} b - \frac{a^{3}}{3} + (b+a) \frac{a^{2}}{2} - \frac{(b+a)^{2}}{4} a \Big) \) \(\displaystyle = \frac{1}{b-a} \Big(\frac{b^{3}-a^{3}}{3} + \frac{b+a}{2} (a^{2}-b^{2}) + \frac{(b+a)^{2}}{4} \ (b-a) \Big) \) \(\displaystyle = \frac{b^{2}+ab+a^{2}}{3} - \frac{(b+a)^{2}}{2} + \frac{(b+a)^{2}}{4} \) \(\displaystyle = \frac{4b^{2}+4ab+4a^{2}-6b^{2}-12ab-6a^{2} + 3b^{2}+6ab + 3a^{2}}{12} \) \(\displaystyle = \frac{b^{2}-2ab+a^{2}}{12} = \frac{(b-a)^{2}}{12} \) Last edited: May 24, 2010

EDIT: \(\displaystyle \sigma^{2} = \frac{1}{b-a} \int^{b}_{a} (x-\mu)^{2} \ dx = \frac{1}{b-a} \int^{b}_{a} \Big(x-\frac{b+a}{2}\Big)^{2} \ dx\) \(\displaystyle = \frac{1}{b-a} \int^{b}_{a} \Big( x^{2} - (b+a) x + \frac{(b+a)^{2}}{4} \Big) \ dx \) \(\displaystyle = \frac{1} {b-a} \Big( \frac{x^{3}}{3} - (b+a) \ \frac{x^{2}}{2} + \frac{(b+a)^{2}}{4} \ x \Big)\Big|^{b}_{a} \) \(\displaystyle = \frac{1}{b-a} \Big(\frac{b^{3}}{3} - (b+a) \frac{b^{2}}{2} + \frac{(b+a)^{2}}{4} b - \frac{a^{3}}{3} + (b+a) \frac{a^{2}}{2} - \frac{(b+a)^{2}}{4} a \Big) \) \(\displaystyle = \frac{1}{b-a} \Big(\frac{b^{3}-a^{3}}{3} + \frac{b+a}{2} (a^{2}-b^{2}) + \frac{(b+a)^{2}}{4} \ (b-a) \Big) \) \(\displaystyle = \frac{b^{2}+ab+a^{2}}{3} - \frac{(b+a)^{2}}{2} + \frac{(b+a)^{2}}{4} \) \(\displaystyle = \frac{4b^{2}+4ab+4a^{2}-6b^{2}-12ab-6a^{2} + 3b^{2}+6ab + 3a^{2}}{12} \) \(\displaystyle = \frac{b^{2}-2ab+a^{2}}{12} = \frac{(b-a)^{2}}{12} \)

D drumist Jan 2010 354 173 May 24, 2010 #5 The mean should be: \(\displaystyle \mu = E[X]= \int_{-\infty}^{\infty} x \, f(x) \, dx = \frac{1}{b-a} \int_a^b x \, dx = \frac{1}{b-a} \cdot \frac{b^2-a^2}{2} = \frac{a+b}{2}\) Reactions: Random Variable

The mean should be: \(\displaystyle \mu = E[X]= \int_{-\infty}^{\infty} x \, f(x) \, dx = \frac{1}{b-a} \int_a^b x \, dx = \frac{1}{b-a} \cdot \frac{b^2-a^2}{2} = \frac{a+b}{2}\)

Random Variable May 2009 959 362 May 24, 2010 #6 drumist said: The mean should be: \(\displaystyle \mu = E[X]= \int_{-\infty}^{\infty} x \, f(x) \, dx = \frac{1}{b-a} \int_a^b x \, dx = \frac{1}{b-a} \cdot \frac{b^2-a^2}{2} = \frac{a+b}{2}\) Click to expand... (Doh) I fixed both of my previous posts. Last edited: May 24, 2010

drumist said: The mean should be: \(\displaystyle \mu = E[X]= \int_{-\infty}^{\infty} x \, f(x) \, dx = \frac{1}{b-a} \int_a^b x \, dx = \frac{1}{b-a} \cdot \frac{b^2-a^2}{2} = \frac{a+b}{2}\) Click to expand... (Doh) I fixed both of my previous posts.

mr fantastic MHF Hall of Fame Dec 2007 16,948 6,768 Zeitgeist May 24, 2010 #7 Nas said: thank you for the help. In order to calculate the variance, do I use the formula where the variance is equal to the integral of (x-mean)^2*f(x)dx? Click to expand... Easier to calculate E(X^2). Then \(\displaystyle Var(X) = E(X^2) - (E(X))^2\).

Nas said: thank you for the help. In order to calculate the variance, do I use the formula where the variance is equal to the integral of (x-mean)^2*f(x)dx? Click to expand... Easier to calculate E(X^2). Then \(\displaystyle Var(X) = E(X^2) - (E(X))^2\).