I'm trying to find the mean out of a cumulative density function (cdf). I found this question but it was no use because I didn't cover the explanation I was expecting. Here's my function: $$ F(x) = \begin{cases} 0, & \text{if x < 0}\\[2ex] x^2, & \text{if 0 $\le$ x < 1/2}\\[2ex] \frac{1}{4}, & \text{if $1/2\le x<3$} \\[2ex] \frac{x-2}{4}, & \text{if $3\le x < 6$}\\[2ex] 1, & \text{if $x\ge 6$} \end{cases} $$

In the question I linked to above, only the interval in the middle was used in the integration. Would anyone mind explaining the intuition behind integrating this function in order to find the mean?

Thank you for your time.

  • $\begingroup$ In the linked post, mean was calculated using the relation $E(X)=\int_0^{\infty}(1-F(x))\,dx$ for a nonnegative RV $X$. If you look carefully, all the pieces were integrated (not just the 'middle' one), then added. $\endgroup$ – StubbornAtom Jun 26 '17 at 18:38

It is said in an answer that the user made an error in the arithmetic.

For your case, I assume the sample space is $[0, 6].$ Here's what we do:

Integrate the CDF as follows:

$$ \int_0 ^6 1 - F(x) \ dx = 6 - \left( \int_0^\frac{1}{2} x^2 \ dx + \int_\frac{1}{2} ^3 \frac{1}{4} \ dx + \int_3 ^6 \frac{x-2}{4} \ dx \right)$$

Can you take it from here?

  • $\begingroup$ the last integral limit shoul be 3 to 6 i guess $\endgroup$ – Upstart Jun 26 '17 at 18:44
  • $\begingroup$ Yes, thank you for catching that. $\endgroup$ – Sean Roberson Jun 26 '17 at 18:44

$F(x) = \int_{0}^{x} f(x) dx\\ \mu = \int_{0}^{6} xf(x) dx$

Integration by parts.

$\mu = xF(x)|_0^6 - \int_{0}^{6} F(x) dx\\ \mu = 6 - \int_{-\infty}^{\infty} F(x) dx\\ \mu = 6 - \int_{0}^{\frac 12} x^2 dx - \int_{\frac 12}^{3} \frac 14 dx- \int_{3}^{6} \frac {x-2}{4} dx$

However you could also say

$f(x) = \frac {d}{dx} F(x) = \begin{cases} 2x &0\le x \le \frac 12\\0 &\frac12 <x \le 3\\\frac {x}{4} &3 <x \le 6\\0&x>6\end{cases}$

$\mu = \int_0^{\frac 12} 2x^2 dx +\int_3^{6} \frac {x}{4} dx $

And, hopefully, those two results are equal.


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