P/1 Actuary Exam Question

Question

I was doing problems and came across this one and was wondering why the $P[1\le x\le 2]$ is $F(2) - \lim\limits_{x\rightarrow1^-} F(x)$ rather than $F(2)-F(1)$? Could someone please explain this for me?

Answer 1

Note that as $x$ approaches $1$ from the left, $F(x)$ approaches $\frac{1}{8}$.

Note also that $F(1)=\frac{1}{4}$: there is a jump. The distribution of our random variable is not continuous. There is a "point mass" of $\frac{1}{4}-\frac{1}{8}$ at $1$, the probability that $X=1$ is $\frac{1}{8}$.

Our required probability is $\Pr(X\le 2)-\Pr(X\lt 1)$. The probability that $X$ is less than $1$ is not $F(1)$. For $F(1)$ is the probability that $X$ is $\le 1$.

Normally, with continuous distributions, we need not distinguish between $\Pr(X\le a)$ and $\Pr(X\lt a)$. For in the case of a continuous distribution, $\Pr(X=a)$ is $0$. But that is not always the case here. And in particular it is not the case at $a=1$.

But always, $\Pr(X\lt a)=\lim_{x\to a^-} F(x)$, so that is what we use here.

Answer 2

If $B \subset A$ then $P(B) - P(A) = P(A\setminus B) \equiv P(A\cap B^c).$ The answer $F(2) - F(1)$ is incorrect since \begin{gather*} F(2) - F(1) = P(\{X \leq 2\}) - P(\{X \leq 1\}) = P(\{X \leq 2\} \cap \{X \leq 1\}^c) = P(1 < X \leq 2),\end{gather*} which is different from $P(1 \leq X \leq 2)$.