# [SOLVED] P.D.F: Continuous random variables

• Mar 15th 2010, 01:32 PM
Quacky
[SOLVED] P.D.F: Continuous random variables
Firstly, I must apologise for the poor notation. I tried to use the \array command but I couldn't get it to work.

1) The random variable Q has P.D.F given by:

$\displaystyle f(q)=kq^3$ if $\displaystyle 0\leq q\leq 2$

$\displaystyle f(q)=0$ otherwise

Find k (answer: $\displaystyle \frac{1}{4}$)

2)The random variable J has probability density function $\displaystyle f(j)=3j^k$ if $\displaystyle 0\leq j\leq 1$

$\displaystyle f(j)=0$ otherwise

I honestly don't know where to start because I missed a lesson and am trying to catch up, but reading and retaining information can be a lot harder than being taught it.
• Mar 15th 2010, 01:42 PM
Plato
Quote:

Originally Posted by Quacky
Firstly, I must apologise for the poor notation. I tried to use the \array command but I couldn't get it to work.

1) The random variable Q has P.D.F given by:

$\displaystyle f(q)=kq^3$ if $\displaystyle 0\leq q\leq 2$

$\displaystyle f(q)=0$ otherwise

Find k (answer: $\displaystyle \frac{1}{4}$)

2)The random variable J has probability density function $\displaystyle f(j)=3j^k$ if $\displaystyle 0\leq j\leq 1$

$\displaystyle f(j)=0$ otherwise

Hint $\displaystyle \int_0^2 {q^3 dq} = 4$
• Mar 15th 2010, 01:50 PM
harish21
Quote:

Originally Posted by Quacky
Firstly, I must apologise for the poor notation. I tried to use the \array command but I couldn't get it to work.

1) The random variable Q has P.D.F given by:

$\displaystyle f(q)=kq^3$ if $\displaystyle 0\leq q\leq 2$

$\displaystyle f(q)=0$ otherwise

Find k (answer: $\displaystyle \frac{1}{4}$)

2)The random variable J has probability density function $\displaystyle f(j)=3j^k$ if $\displaystyle 0\leq j\leq 1$

$\displaystyle f(j)=0$ otherwise

I honestly don't know where to start because I missed a lesson and am trying to catch up, but reading and retaining information can be a lot harder than being taught it.

1) If $\displaystyle f(q)=kq^3$ is the pdf of the random variable, then we know that

$\displaystyle \int_0^2 {kq^3 dq} = 1$

or, $\displaystyle k [\frac{q^4}{4}]_0^2 = 1$

or, $\displaystyle k [\frac{2^4}{4} - \frac{0}{4}] = 1$

or, $\displaystyle k \frac{16}{4} = 1$

0r, $\displaystyle k = \frac{1}{4}$

Try doing the same for your second problem. Show If you come up with any problems. By the way, are you supposed to find j or k in your second question? I assume you are supposed to find k.
• Mar 15th 2010, 01:52 PM
Quacky
Thankyou Plato and Harish, I understand now. I shall attempt 2) by myself.
• Mar 15th 2010, 02:10 PM
Quacky
After a minor integration error, I've solved part 2). Thanks again