single varibale cummulative dist. func.

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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.