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I'm a little confused with the next excercise:

Let $F$ a probability distribution given by $$F(x)=\left\{\begin{array}{cc}0&\mbox{if}x<4\\1/5&\mbox{if }4\le x<6\\7/10&\mbox{if }6\le x<8\\23/30&\mbox{if } 8\le x<10\\1&\mbox{if }10\le x\end{array}\right.$$

Calculate $f(6)$.

I'm sure that $F$ does not have density function, but some others tell me that $F$ is discrete. How do you try this problem?

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    $\begingroup$ $F$ is piecewise constant, so the distribution is discrete. Presumably "$f$" refers to a probability mass function (PMF) rather than a density function, and $f(6)$ is asking for the probability assigned to the value $6$. $\endgroup$
    – angryavian
    Oct 8 at 4:20
  • $\begingroup$ $\displaystyle\operatorname{F}'\left(x\right) = {6 \over 5}\delta\left(x - 4\right) + {7 \over 10 }\delta\left(x - 6\right) + {23 \over 30}\delta\left(x - 8\right) + \delta\left(x - 10\right)$ $\endgroup$ Oct 8 at 5:10
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$F$ is a cumulative distribution function - describing how probability mass accumulates with increasing $x$ .

Here we can see that the probability mass does not accumulate inside the parts (it remains constant), but rather has a massive step increase between the parts.   Therefore this CDF does describe a discrete distribution.

As such will not have a probability density function. Rather it will have a probability mass function.

Proceed under the premise that $f$ is intended to be that pmf .

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