# single varibale cummulative dist. func.

• April 12th 2008, 09:04 PM
lllll
single varibale cummulative dist. func.
$f(x) = \left\{ \begin{array}{rcl}
\frac{3}{4} x^2(1-x^2) & \mbox{for} & -1 < x < 1 \\
0 & \mbox{for} & \mbox{other}
\end{array}\right.$

find the cumulative distribution function for $P(X \leq x)$

so far I get

$f(x) = \left\{ \begin{array}{rcl}
0 & \mbox{for} & x \leq -1 \\
\frac{3}{4}x - \frac{1}{4}x^3 & \mbox{for} & -1 < x < 1 \\
1 & \mbox{for} & x \geq 1
\end{array}\right.$

but the given solution is:

$f(x) = \left\{ \begin{array}{rcl}
0 & \mbox{for} & x \leq -1 \\
\frac{3}{4}x - \frac{1}{4}x^3 + \color{red} \frac{2}{3} & \mbox{for} & -1 < x < 1 \\
1 & \mbox{for} & x \geq 1
\end{array}\right.$

where those the $\frac{2}{3}$ come from?
• April 12th 2008, 09:32 PM
mr fantastic
Quote:

Originally Posted by lllll
$f(x) = \left\{ \begin{array}{rcl}
\frac{3}{4} x^2(1-x^2) & \mbox{for} & -1 < x < 1 \\
0 & \mbox{for} & \mbox{other}
\end{array}\right.$
Mr F asks: Is this the pdf? If so, why do you have f(x) below (*)?

find the cumulative distribution function for $P(X \leq x)$

so far I get

$f(x) = \left\{ \begin{array}{rcl}
0 & \mbox{for} & x \leq -1 \\
\frac{3}{4}x - \frac{1}{4}x^3 & \mbox{for} & -1 < x < 1 \\
1 & \mbox{for} & x \geq 1
\end{array}\right.$
* ??

but the given solution is:

$f(x) = \left\{ \begin{array}{rcl}
0 & \mbox{for} & x \leq -1 \\
\frac{3}{4}x - \frac{1}{4}x^3 + \color{red} \frac{2}{3} & \mbox{for} & -1 < x < 1 \\
1 & \mbox{for} & x \geq 1
\end{array}\right.$

where those the $\frac{2}{3}$ come from?

Regardless of your answer, note that to get the cdf you integrate the pdf from -1 to x .....

Substituting -1 into the antiderivative of the pdf will account for any constant you get in the cdf .....
• April 12th 2008, 10:19 PM
lllll
those are typos and I didn't copy it accordingly, what's shown for the cdf is:

$F(x) = \left\{ \begin{array}{rcl}
0 & \mbox{for} & x \leq -1\\
\int^{x}_{1} \frac{3}{4}(1-u^2) \ du & \mbox{for} & -1 < x < 1 \\
1 & \mbox{for} & x \geq 1
\end{array}\right.$

$F(x) = \left\{ \begin{array}{rcl}
0 & \mbox{for} & x \leq -1\\
\frac{3}{4}x -\frac{1}{4}x^3 + \color{red} \frac{2}{3}& \mbox{for} & -1 < x < 1 \\
1 & \mbox{for} & x \geq 1
\end{array}\right.$

but after integrating out you get:

$\frac{3}{4}u-\frac{1}{4}u^3 \bigg{|}^{x}_{1} = \frac{3}{4}x-\frac{1}{4}x^3 - \left( \frac{3}{4}(-1) - \frac{1}{4}(-1)^3 \right) = \frac{3}{4}x-\frac{1}{4}x^3 +\frac{1}{2}$

so either I'm missing something or there is in fact an error in the solution...
• April 12th 2008, 11:11 PM
CaptainBlack
Quote:

Originally Posted by lllll
those are typos and I didn't copy it accordingly, what's shown for the cdf is:

$F(x) = \left\{ \begin{array}{rcl}
0 & \mbox{for} & x \leq -1\\
\int^{x}_{1} \frac{3}{4}(1-u^2) \ du & \mbox{for} & -1 < x < 1 \\
1 & \mbox{for} & x \geq 1
\end{array}\right.$

$F(x) = \left\{ \begin{array}{rcl}
0 & \mbox{for} & x \leq -1\\
\frac{3}{4}x -\frac{1}{4}x^3 + \color{red} \frac{2}{3}& \mbox{for} & -1 < x < 1 \\
1 & \mbox{for} & x \geq 1
\end{array}\right.$

but after integrating out you get:

$\frac{3}{4}u-\frac{1}{4}u^3 \bigg{|}^{x}_{1} = \frac{3}{4}x-\frac{1}{4}x^3 - \left( \frac{3}{4}(-1) - \frac{1}{4}(-1)^3 \right) = \frac{3}{4}x-\frac{1}{4}x^3 +\frac{1}{2}$

so either I'm missing something or there is in fact an error in the solution...

What happed to the other $x^2$ in the density?

RonL
• April 12th 2008, 11:19 PM
mr fantastic
Quote:

Originally Posted by lllll
those are typos and I didn't copy it accordingly, what's shown for the cdf is:

$F(x) = \left\{ \begin{array}{rcl}
0 & \mbox{for} & x \leq -1\\
\int^{x}_{1} \frac{3}{4}(1-u^2) \ du & \mbox{for} & -1 < x < 1 \\
1 & \mbox{for} & x \geq 1
\end{array}\right.$

$F(x) = \left\{ \begin{array}{rcl}
0 & \mbox{for} & x \leq -1\\
\frac{3}{4}x -\frac{1}{4}x^3 + \color{red} \frac{2}{3}& \mbox{for} & -1 < x < 1 \\
1 & \mbox{for} & x \geq 1
\end{array}\right.$

but after integrating out you get:

$\frac{3}{4}u-\frac{1}{4}u^3 \bigg{|}^{x}_{1} = \frac{3}{4}x-\frac{1}{4}x^3 - \left( \frac{3}{4}(-1) - \frac{1}{4}(-1)^3 \right) = \frac{3}{4}x-\frac{1}{4}x^3 +\frac{1}{2}$

so either I'm missing something or there is in fact an error in the solution...

$\int^{x}_{1} \frac{3}{4}(1-u^2) \ du = \frac{3}{4}x - \frac{1}{4}x^3 + \frac{1}{2}$ is correct ...... But the pdf and hence integral is up in the air I think ....
• April 13th 2008, 01:03 AM
CaptainBlack
Quote:

Originally Posted by mr fantastic
$\int^{x}_{1} \frac{3}{4}(1-u^2) \ du = \frac{3}{4}x - \frac{1}{4}x^3 + \frac{1}{2}$ is correct ...... But the pdf and hence integral is up in the air I think ....

Well I don't see any reason for us to worry about the algebra so I let Maxima do the work:

Code:

(%i5) integrate((3/4)*(1-z^2), z, -1, x); (%o5) (3*(2/3-(x^3-3*x)/3))/4 (%i6) radcan(%); (%o6) -(x^3-3*x-2)/4 (%i7)
and as this is $1$ at $x=1$, we have that $(3/4)(1-x^2)$ for $x \in (-1,1)$ and zero otherwise is a density, but the $(3/4)x^2 (1-x^2)$ of the original post is not (it integrates up to $1/5$)

RonL

(note Maxima unlike Mathematica is freeware - and does not just produce plain ASCII output as above either)
• April 13th 2008, 04:13 AM
mr fantastic
Quote:

Originally Posted by CaptainBlack
Well I don't see any reason for us to worry about the algebra so I let Maxima do the work:

Code:

(%i5) integrate((3/4)*(1-z^2), z, -1, x);   (%o5) (3*(2/3-(x^3-3*x)/3))/4   (%i6) radcan(%);   (%o6) -(x^3-3*x-2)/4   (%i7)
and as this is $1$ at $x=1$, we have that $(3/4)(1-x^2)$ for $x \in (-1,1)$ and zero otherwise is a density, but the $(3/4)x^2 (1-x^2)$ of the original post is not (it integrates up to $1/5$)

RonL

(note Maxima unlike Mathematica is freeware - and does not just produce plain ASCII output as above either)

Good thinking, 99.