# Thread:

1. ## 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.

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

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

2. evaluate if:

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

3. 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.)

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

So: k=3/2.

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

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...

6. 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

7. Answer:

f(x) < 0.

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

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

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

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.

9. 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...

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

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

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.

11. 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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