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I am given the following distribution function of the r.v. X (first is for x < -2)

I need to find the pmf for $p(0), p(1)$,

There's something that I don't understand in the computation of the pmf using the cdf. I know that we can write that:

Therefore I conclude that

  • $p(0) = 0.5 - 0.2 = 0.3$
  • $p(1) = 0.5 - 0.2 = 0.3$
  • $ = p(p(-2) + p(0)) = p(0.5) = 0.5 - 0.2 = 0.3$

It happens that only $p(0)$ is correct and the other two are wrong. I see that that the ones that are wrong, for example p(1), should be equal to zero because "it is not located on a jump" and therefore cannot be used in the aforementioned formula.

Do you have an another explanation than my pretty poor observation ?

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1 Answers 1

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In general:$$p(x):=P(X=x)=P(X\leq x)-P(X

So that: $$p(0)=F(0)-\lim_{y\uparrow 0}F\left(y\right)=0.5-0.2=0.3$$ and:$$p(1)=F(1)-\lim_{y\uparrow 1}F\left(y\right)=0.5-0.5=0$$ and:$$p(\Pr(X\leq0)=p(F(0))=p(0.5)=F(0.5)-\lim_{y\uparrow 0.5}F\left(y\right)=0.5-0.5=0$$