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.