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I have the following probability density function (pdf):

$f(x)=\{0$ if $x<= 11$

$x-11$ if $11\leq x \leq 12$

$13-x$ if $12\leq x \leq 13 \}$

My aim is to find the distribution function $F(x)$ where the integral from $-\infty$ to $+\infty$ of the pdf $f(x)$ equals one. I know I have to integrate the pdf over the various sub-intervals indicated above, however, I get something that is not correct. Could somebody please explain me step by step how to do it? I apologize for my bad typing of the problem, I hope you can understand. Thank you in advance

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    $\begingroup$ The $F(x)$ in your 5th line, is it a cumulative distribution function ? $\endgroup$ Commented Nov 16, 2018 at 11:55
  • $\begingroup$ Yes it is also a cdf $\endgroup$ Commented Nov 16, 2018 at 13:04

1 Answer 1

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Since the value of the pdf before $11$ is $0$, $F(x)$ will also be $0$ before $11$. After $11$, we integrate pdf from $11$ to $12$ separately, and $12$ to $13$ separately as they have different pdfs.

From $11$ to $12$ ($11\leq y\leq 12$)

$$F(y) = \int_{11}^{y}(x - 11)dx = \frac{y^2}{2} - 11y + \frac{121}{2} $$

For $12$ to $13$ ($12\leq y\leq 13$), we'll need the value of $F(12)$ as this will be added to the integral. So putting $y = 12$ in the above integral, we get $F(12) = 0.5$

Now for $12$ to $13$ ($12\leq y\leq 13$)

$$F(y) = 0.5 + \int_{12}^{y}(13 - x)dx = 13y - \frac{y^2}{2} - 83.5$$

After $13$, i.e. $y\geq 13$

$$F(y) = 1$$

as can be seen by putting $y = 13$ in the second equation.

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  • $\begingroup$ Thank you Sauhard! However, the solution that I have reads: $0$ for $x\leq 11$; $\frac{1}{2}(x-11)^2$ for $11 \leq x \leq 12$; $1-\frac{1}{2}(13-x)^2$ for $12 \leq x \leq 13$ and $1$ for $x \geq 12$ $\endgroup$ Commented Nov 16, 2018 at 13:05
  • $\begingroup$ You can factorise my equations so that they fit your way of solution. For $11\leq y \leq 12$ , rewrite my equation of $F(y) = \frac{y^2}{2} - 11y + \frac{121}{2} = \frac{1}{2}(y^2 - 22y + 121) = \frac{1}{2}(y-11)^2$. For $12 \leq y \leq 13$, rewrite $F(y) = 13y - \frac{y^2}{2} - 83.5 = 1 +13y - \frac{y^2}{2} - 84.5 = 1 - \frac{1}{2}(y^2 - 26y + 169) = 1 - \frac{1}{2}(13 - y)^2$ $\endgroup$ Commented Nov 16, 2018 at 13:21

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