# Continuous random variables – probability density functions

• Nov 4th 2008, 07:18 PM
geton
Continuous random variables – probability density functions
The following could represent a probability density function (p.d.f.) or not. If could, find the value of k.

$\displaystyle f(x) = \begin{cases} k(x + x^2), & \text {$-1 \le x \le 1,$} \\ 0, & \text {otherwise.} \end{cases}$

How could I know it represent a p.d.f or not?
• Nov 4th 2008, 07:28 PM
Black Kawairothlite
evaluate if:

$\displaystyle \int_{-\infty}^{\infty} f(x)dx=1$
• Nov 4th 2008, 08:24 PM
geton
Well, my problem is not to finding k. I’ve confusion to recognize is it p.d.f or not.

If I try on this way,

f(-0.5) = k(0.75)

f(1) = 2k

I didn’t understand how the above info/working or else will show

f(x) < 0. So it could not be a p.d.f. (As my book’s answer.)
• Nov 4th 2008, 08:29 PM
geton
If $\displaystyle \int_{-1}^{1} f(x)dx=1$

So: k=3/2.

:confused::confused::confused:
• Nov 4th 2008, 09:15 PM
Black Kawairothlite
$\displaystyle \int_{-\infty}^{\infty} f(x)dx=1$

you gotta "chop" the integral in three parts:

$\displaystyle \int_{-\infty}^{-1} 0dx+\int_{-1}^{1} k(x + x^2)dx+\int_{1}^{\infty} 0dx=1$

continue...
• Nov 4th 2008, 09:19 PM
Black Kawairothlite
Quote:

Originally Posted by geton
If $\displaystyle \int_{-1}^{1} f(x)dx=1$

So: k=3/2.

:confused::confused::confused:

that's the right way can't say if that's the answer gotta do it but im gonna sleep atm
• Nov 4th 2008, 10:24 PM
geton

f(x) < 0.

Therefore f(x) cannot be a probability density function.
• Nov 4th 2008, 10:26 PM
mr fantastic
Quote:

Originally Posted by geton
The following could represent a probability density function (p.d.f.) or not. If could, find the value of k.

$\displaystyle f(x) = \begin{cases} k(x + x^2), & \text {$-1 \le x \le 1,$} \\ 0, & \text {otherwise.} \end{cases}$

How could I know it represent a p.d.f or not?

$\displaystyle f(x) = k(x + x^2)$ can only represent a pdf if it's non-zero for all values of x. This is clearly NOT the case therefore it's cannot represent a pdf.
• Nov 4th 2008, 11:28 PM
geton
Quote:

Originally Posted by mr fantastic
$\displaystyle f(x) = k(x + x^2)$ can only represent a pdf if it's non-zero for all values of x. This is clearly NOT the case therefore it's cannot represent a pdf.

Why I'm trying to prove f(x) < 0? Totally time wasting...
• Nov 5th 2008, 05:22 AM
awkward
Quote:

Originally Posted by geton
The following could represent a probability density function (p.d.f.) or not. If could, find the value of k.

$\displaystyle f(x) = \begin{cases} k(x + x^2), & \text {$-1 \le x \le 1,$} \\ 0, & \text {otherwise.} \end{cases}$

How could I know it represent a p.d.f or not?

Hi geton,

One of the requirements for a pdf is that $\displaystyle f(x) \geq 0$ for all values of x. Now

$\displaystyle x + x^2 < 0 \text{ for } -1 < x < 0$

and

$\displaystyle x + x^2 > 0 \text{ for } 0 < x < 1$,

so if $\displaystyle k > 0$ then $\displaystyle f(x) \geq 0$ fails for $\displaystyle -1 < x < 0$, and if $\displaystyle k < 0$ then $\displaystyle f(x) \geq 0$ fails for $\displaystyle 0 < x < 1$, so no value of $\displaystyle k \neq 0$ will make f suitable for a pdf.

The only remaining possibility is $\displaystyle k = 0$ but then $\displaystyle \int f(x) dx = 0$, so that won't work either.
• Nov 6th 2008, 02:17 AM
geton
Quote:

Originally Posted by awkward
Hi geton,

One of the requirements for a pdf is that $\displaystyle f(x) \geq 0$ for all values of x. Now

$\displaystyle x + x^2 < 0 \text{ for } -1 < x < 0$

and

$\displaystyle x + x^2 > 0 \text{ for } 0 < x < 1$,

so if $\displaystyle k > 0$ then $\displaystyle f(x) \geq 0$ fails for $\displaystyle -1 < x < 0$, and if $\displaystyle k < 0$ then $\displaystyle f(x) \geq 0$ fails for $\displaystyle 0 < x < 1$, so no value of $\displaystyle k \neq 0$ will make f suitable for a pdf.

The only remaining possibility is $\displaystyle k = 0$ but then $\displaystyle \int f(x) dx = 0$, so that won't work either.

Thank you so much. I really appreciate it. :)