# What is the difference between a continuous distribution function and a cumulative distribution function?

My book calls them "CDF" for both and I cannot tell the difference.

For example I have a problem.

A non-negative integer valued random variable X has a CDF of the form $F(x) = 1 - (1/2)^{x-1}$ for $x = 0,1,2,...$ and zero if $x<0$

What is the pdf of X?

What is $P(10 < X \leq 20)$

My book says the pdf is $(1/2)^{x-1}$ and I have no idea why

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"Continuous" refers to the values your random variable can take: a discrete set like the integers, or a continuum like the real numbers? "Cumulative" means that the density function measures $P(X\leq x)$, the probability that $X$ is in the range of values up to and including $x$ (like how a cumulative exam covers everything you've learned up till now).

To recover the probability density function (pdf) from the cumulative density function, you either need to differentiate (if the random variable has a continuous distribution) or take a difference (if the random variable is discrete).

In particular, in your example, the distribution looks discrete since values for the cumulative density function are only given for integer values of $x$. (You can tell it's "cumulative" because the values increase to $1$ instead of summing/integrating to $1$.) So to find the probability density function $P(X=x)$, you can find it by taking the difference in two neighboring values of $F$: \begin{align*} P(X=x) &= P(x-1 < X \leq x)\\ &= P(X\leq x) - P(X\leq x-1)\\ &= F(x) - F(x-1)\\ &= \left(1-\frac{1}{2^{x-1}}\right)-\left(1-\frac{1}{2^{x-2}}\right)\\ &= \frac{1}{2^{x-2}} - \frac{1}{2^{x-1}}\\ &= \frac{1}{2^{x-1}} \end{align*} In general, to find $P(a< X\leq b)$ from the cumulative density function, just note that $X\leq a$ and $a< X\leq b$ are disjoint events whose union is $X\leq b$, so $$P(X\leq a) + P(a < X\leq b) = P(X\leq b)$$ $$F(a) + P(a<X\leq b) = F(b)$$ $$P(a< X\leq b) = F(b) - F(a).$$

I'll leave it to you to find the value of $P(10<X\leq 20)$ in this example.

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