# Continuous Random variable

• Nov 10th 2010, 01:24 PM
axa121
Continuous Random variable
Hi, I'm unsure where to go with this question:

Let $f(a,b) (x) \begin{cases} = { \left ( \frac{bx}{a} \right ) for \in [0,a] \\ {0} otherwise \end{cases}$

If a > 0 for what values of b > 0 is f(a,b) the pdf of a continuous variable. Also for what values of a and b is the E(X) = 1/3

So what I'm doing is $\int_{a}^{0} {bx/a} \, dx$ and then I get $-(a b)/2$ . I was thinking to make this equal to 1, but don't no what to do next.
• Nov 10th 2010, 01:29 PM
mr fantastic
Quote:

Originally Posted by axa121
Hi, I'm unsure where to go with this question:

Let $f(a,b) (x) \begin{cases} = { \left ( \frac{bx}{a} \right ) for \in [0,a] \\ {0} otherwise \end{cases}$

If a > 0 for what values of b > 0 is f(a,b) the pdf of a continuous variable. Also for what values of a and b is the E(X) = 1/3

So what I'm doing is $\int_{a}^{0} {bx/a} \, dx$ and then I get $-(a b)/2$ . I was thinking to make this equal to 1, but don't no what to do next.

You require:

1. $\displaystyle \frac{bx}{a} \geq 0$ for $0 \leq x \leq a$.

2. $\displaystyle \int_0^a \frac{bx}{a} \, dx = 1$.

And for the third part, you require $\displaystyle \int_0^a \frac{bx^2}{a} \, dx = \frac{1}{3}$.
• Nov 10th 2010, 03:11 PM
axa121
Quote:

Originally Posted by mr fantastic
You require:

1. $\displaystyle \frac{bx}{a} \geq 0$ for $0 \leq x \leq a$.

2. $\displaystyle \int_0^a \frac{bx}{a} \, dx = 1$.

And for the third part, you require [tex]\displaystyle \int_0^a \frac{bx^2}{a} \, dx = \frac{1}{3}[\math].

I got a = 1 and b = 1 for the expected value to be 1/3

But do I need to solve the inqualities for the first bit. After integrating I have ab/2 = 1, but the first equation has [tex] \displaystyle \frac{bx}{a} \geq 0 [\math] which has an x in it.
• Nov 11th 2010, 01:58 PM
axa121
Im still stuck, after integrating for the first part I get ab = 2 how can I solve this?
• Nov 11th 2010, 03:36 PM
mr fantastic
Quote:

Originally Posted by axa121
Im still stuck, after integrating for the first part I get ab = 2 how can I solve this?

You can't. All you can do is get a in terms of b. If you are expected to simultaneously satisfy the expected value being equal to 1/3, then you solve the following simultaneously:

ab = 2

ba^2 = 1

(I have no idea where your answers in post #3 came from but they are not correct).
• Nov 11th 2010, 03:45 PM
axa121
Right so a = 1/2 and b = 4, this gives E(X) = 1/3. But the first part asks for what values of b > 0 is f(a,b)(x) the pdf of a continuous random variable.
• Nov 11th 2010, 04:00 PM
mr fantastic
Quote:

Originally Posted by axa121
Right so a = 1/2 and b = 4, this gives E(X) = 1/3. But the first part asks for what values of b > 0 is f(a,b)(x) the pdf of a continuous random variable.

b = 2/a.
• Nov 11th 2010, 04:05 PM
axa121
I see, I was looking for one or two specific values. Thank you very much.