I understand that to the expected value for the random variable X but with the cumulative distribution function (shown below), I am unable to differentiate it to find the probability density function to solve for the expected value.

$$F_X(x)=\begin{cases}0 & for \ \ x < 0.8\ \\0.2 & for \ \ 0.8 ≤ x < 1.2\ \\0.45 & for \ \ 1.2 ≤ x < 1.8\ \\0.75 & for \ \ 1.8≤ x < 2.5\ \\0.9 & for \ \ 2.5≤ x < 4\ \\ 1 & for \ \ x≥4 \end{cases}$$

Is there another way for me to find the expected value of X without having to find the probability density function?

I would greatly appreciate any help that can point me in the right direction for this question.Thank you!



First note that there is an error in your question, since the first condition is $x<0.8$ and the second condition is $0.7 \le x < 1.2$. They should be disjoint.

Note that from the CDF, we are able to compute the pmf, for example,

$$P(X=1.2)= P(X \le 1.2)-P(X < 1.2)=0.45-0.2=0.25$$

After you compute the pmf, you can compute the expected value.

  • $\begingroup$ Are not you assuming that $X$ is a discrete r.v.? If it is continuous, $P(X=1.2)=0$. $\endgroup$ – farruhota Oct 6 at 9:51
  • $\begingroup$ yes, the function is discontinous at those points. $\endgroup$ – Siong Thye Goh Oct 6 at 10:00

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