0
$\begingroup$

I am working on a problem and am a bit stuck. It is finding a CDF from a continuous distribution.

The problem:

$$ f(x) = \cases{.005(20-x) & for 0$\lt$x$\lt$20\cr 0 & otherwise\cr} $$

Find the CDF for the random variable.

What I have done so far:

$F_x(x) = P(X\le x)$ or $F_x(a) = P(X\le a)$

(The integral of the pdf)

if x is continuous then $F_x(a)$ = $\int_{-\infty}^ap(y)d(y)$ where p is the density

I am unsure how to find this and what $a$ should we use?

$\endgroup$
2
$\begingroup$

Your density function $f$ has finite support, so you should take appropriate care for your integral's limits. Specifically, for $x \leq 0$, your CDF will be

$$ F_{X}(x) = \int_{-\infty}^0 f(s) \text{d}s = 0 $$

In the same spirit, since integrating a PDF over its support sums to $1$, you will get that for $x \geq 20$, your CDF will satisfy

$$ F_X(x) = \int_{-\infty}^{x} f(s) \text{d} s = \int_{0}^{20} f(s) \text{d}s = 1 $$

Note that the integral's lower limit starts from $0$, since $f(s)$ is $0$ for $s \leq 0$. Now, for any $x \in (0, 20)$ you must calculate the integral explicitly. You obtain

$$ F_X(x) = \int_{-\infty}^x f(s) \text{d}s = \int_0^x 0.005 (20 - s) \text{d} s = 0.005 \left(20 x - \int_0^x s \text{d} s\right) \\ = 0.005 \left(20x - \frac{x^2}{2} \right), \; 0 < x < 20. $$

As a sanity check, you can plot it on Wolfram Alpha to make sure it looks like a CDF.

$\endgroup$
  • $\begingroup$ So just as clarification, we do not get a finite number at the end of our calculation for the CDF, just the function itself. $\endgroup$ – Ethan Nov 5 '18 at 19:05
  • $\begingroup$ @Ethan: Yes, that is correct. $\endgroup$ – VHarisop Nov 5 '18 at 19:07
  • $\begingroup$ Thank you so much $\endgroup$ – Ethan Nov 5 '18 at 19:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.