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?


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.

  • $\begingroup$ 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

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