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Suppose that the cumulative distribution function of a random variable X is given by

$ F(a) = \begin{cases} 0,& a < 0 \\ 1/5, & 0 \leq a < 2 \\ 2/5, & 2 \leq a < 4 \\ 1, & a \geq 4 \end{cases} $

Find the probability mass function of X?

My reasoning is as follows: The cdf is discontinuous at the points 0, 2, and 4. Between these $F'(a)$ is defined and $=0$, hence the pmf needs definition only at these points. But how do we get the probabilities at a = 0, 2, 4?

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$P(X\leq0)=\frac15$ and from $P(X\leq a)=0$ for each $a<0$ it follows that $P(X<0)=0$.

Then $$P(X=0)=P(X\leq0)-P(X<0)=\frac15-0=\frac15$$

The others (at $2$ and $4$) can be found likewise.

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HINT:

$$F'(4)=F(a>4)-F(2\le a<4)$$ $$F'(2)=F(2\le a<4)-F(0\le a<2)$$ $$F'(0)=F(0\le a<2)-F(a<0)$$

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  • $\begingroup$ You are right, but I didn't understand why? $\endgroup$ – buzaku Nov 8 '15 at 13:18
  • $\begingroup$ Cumulative distribution function $F(a)$ is the sum of the probabilities from $x=0$ to $x=a$. So the pmf $F'(a)$ is obtained by taking the respective differences for the points where the function is discontinuous....Is it clear?? $\endgroup$ – SchrodingersCat Nov 8 '15 at 13:21
  • $\begingroup$ What is the meaning of e.g. $F(a>4)$ where $F$ is a function? Yes, I understand what you mean, but is this an acknowledged notation for that? $\endgroup$ – drhab Nov 8 '15 at 13:47
  • $\begingroup$ @drhab I don't know for sure. I used for proper understanding only. Better notation would have been $F(a)\big|_{a>4}$. $\endgroup$ – SchrodingersCat Nov 8 '15 at 13:51

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