If I have the pmf of a random variable X such that: $$ f(x) = \begin{cases} 1, & \text{if } x=0 \\ 0, & \text{elsewhere} \end{cases} $$ Would the cdf just be the same as the pmf. Such that: $$ F(x) = \begin{cases} 1, & \text{if } x=0 \\ 0, & \text{elsewhere} \end{cases} $$ where $f(x)$ is the pmf, and $F(x)$ is the cdf. I figured this was to simple to actually be true which is why I am asking. Because in my class and in the textbook it only ever does examples with multiple cases or a bigger range for its values.

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    $\begingroup$ The "c" in "cdf" stands for "cumulative", so no that's not how it works. $\endgroup$ – Michael Hardy Oct 2 '17 at 0:25

No. The cdf $F(x)$ "cumulates" all pmf values $f(t)$ for $t\le x$ :

$$F(x)=\sum_{t\le x}f(t)= \begin{cases} 0, & \text{ if } x<0\\ 1, & \text{ if } x\ge 0 \end{cases}$$

  • $\begingroup$ Oh ok thank you for your help, I know thats what we did for pmf's with multiple values, and more then one x values, so I wasn't 100% sure if it would still be the same because there was only one x value and f(x) value. $\endgroup$ – Robert Oct 2 '17 at 0:27

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