Given $f(y) = \lambda e^{-\lambda y}$ , for $y>0$, find $P(Y>s|Y>t)$

Question

Random variable y with pdf as follows.

Given $f(y) = \lambda e^{-\lambda y}$ , for $y>0$, find $P(Y>s|Y>t)$.

I am having difficulty computing this. I have used bayes and integrated to get the CDF and then used $1-P(y<t)$ and $1-P(y<s)$ and cannot get the correct answer. How do I solve this?

Answer 1

Here's a hint.

You don't need Bayes' theorem. Assume that $s > t \geq 0$.

Then $$P(Y > s \mid Y > t) = \dfrac{P(Y > s \cap Y > t)}{P(Y > t)} = \dfrac{P(Y > s)}{P(Y > t)}$$ Can you find these two quantities?

Here's why $P(Y > s \cap Y > t) = P(Y > s)$:

The interval in which the green and red lines both appear above is wherever $Y > s$.

Answer 2

If $t<s$, then $\Pr(Y>s|Y>t)=0$. Otherwise, with $s\geq t$, $$ \Pr(Y>s|Y\geq t)=\frac{\Pr(Y>s\text{ and }Y>t)}{\Pr(Y\geq t)}=\frac{\Pr(Y>s)}{\Pr(Y\geq t)}=\frac{\int_s^\infty\lambda\exp(-\lambda y)dy}{\int_t^\infty\lambda\exp(-\lambda y)dy}. $$ For any $\tau\geq 0$, note that $\int_\tau^\infty\lambda\exp(-\lambda y)dy=\exp(-\lambda\tau)$. Can you complete the solution?