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An investment firm offers its customers municipal bonds that mature after varying numbers of years. Given that the cumulative distribution of T, the number of years to maturity for a randomly selected bond is:

$$F(t) = \begin{cases} 0, & t<1 \\ \frac 14, & 1\le t<3 \\ \frac 12, &3\le t<5 \\ \frac34, & 5\le t<7 \\ 1, &t\ge 7\end{cases}$$

Find $P(T=5)$, $P(T>3)$, and $P(1.4 < T <6)$.

I am unsure how to use the cdf for $P(t=5)$ because it is right between the given ranges. Is the answer $0+1/4+1/2 = 3/4$?

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2 Answers

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By defintion, the cdf $F$ of a random variable $X$ satisfies $F(t) = P(X\leq t)$. (Based on your post, it sounds like you might have $F$ and $P$ reversed in that equation.) Thus $$P(T = 5) = P(T \leq 5) - P(T < 5) = \frac{3}{4}- \frac{1}{2} = \frac{1}{4},$$ since $P(T \leq 5 - \epsilon) = F(5 - \epsilon) = \frac{1}{2}$ for small $\epsilon > 0$. For the latter two statements, use the fact that $P(X > t) = 1 - P(X\leq t)$ and $P(s < X < t) = P(X < t) - P(X \leq s)$.

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(a) $P(T = 5) = F(5) − F(4) = \frac34 − \frac12 = \frac14$

(b) $P(T > 3) = 1 − F(3) = 1 − \frac12 = \frac12$

(c) $P(1.4 < T < 6) = F(6) − F(1.4) = \frac34 − \frac14 = \frac12$

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