Continuous random variables – probability density functions

**geton** · November 4th 2008, 07:18 PM

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

$ 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?

**Black Kawairothlite** · November 4th 2008, 07:28 PM

evaluate if:

$\int_{-\infty}^{\infty} f(x)dx=1$

**geton** · November 4th 2008, 08:24 PM

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.)

**geton** · November 4th 2008, 08:29 PM

If $\int_{-1}^{1} f(x)dx=1$

So: k=3/2.

**Black Kawairothlite** · November 4th 2008, 09:15 PM

$\int_{-\infty}^{\infty} f(x)dx=1$

you gotta "chop" the integral in three parts:

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

continue...

**Black Kawairothlite** · November 4th 2008, 09:19 PM

Originally Posted by geton

If $\int_{-1}^{1} f(x)dx=1$

So: k=3/2.

that's the right way can't say if that's the answer gotta do it but im gonna sleep atm

**geton** · November 4th 2008, 10:24 PM

Answer:

f(x) < 0.

Therefore f(x) cannot be a probability density function.

**mr fantastic** · November 4th 2008, 10:26 PM

Originally Posted by geton

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

$ 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?

$f(x) = k(x + x^2)$ 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.

**geton** · November 4th 2008, 11:28 PM

Originally Posted by mr fantastic

$f(x) = k(x + x^2)$ 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...

**awkward** · November 5th 2008, 05:22 AM

Originally Posted by geton

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

$ 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 $f(x) \geq 0$ for all values of x. Now

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

and

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

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

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

**geton** · November 6th 2008, 02:17 AM

Originally Posted by awkward

Hi geton,

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

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

and

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

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

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

Thank you so much. I really appreciate it.

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