3

Given $X \sim exp(\lambda)$ and a constant $c \ge 0$. Let $Y=\min(X,c)$. I would like to find $E[Y]$ and tried looking for possible approaches in link1 and link2.

However, the accepted answers in both URLs give very different results. For example, according to the answer in link1

\begin{align*} \mathsf E(Y) & = \mathsf E(Y\mid X\leq c)\,\mathsf P(X\leq c)+\mathsf E(Y\mid X>c)\,\mathsf P(X> c) \\ & = \mathsf E(X\mid X\leq c)\,\mathsf P(X\leq c) +\mathsf E(c\mid X>c)\,\mathsf P(X>c) \\ & = \mathsf E(X\mid X\leq c)\,\mathsf P(X\leq c) +c\,\mathsf P(X>c) \\ & = (1-e^{-\lambda c})\int_0^c t\lambda e^{-\lambda t}dt + ce^{-\lambda c} \end{align*}

and according to the accepted answer in link2: \begin{align*} E(Y)&= \int_0^\infty P( Y > t) dt \\ &= \int_0^\infty P( X > t, c > t) dt \\ &= \int_0^\infty P( X > t)1\{ c > t\} dt \\ &= \int_0^c P(X > t)dt \\ &= \int_0^c e^{-\lambda t}dt \end{align*}

What even confused me more is another direct solution to this problem shown in the second answer in link1.

Which one is correct? Honestly I am not sure how to start with this problem, but the first answer seems to make more sense to me.

user36706
  • 79
  • 4

2 Answers2

2

The first one is wrong. The first term there should be $\int_0^{c} t\lambda e^{-\lambda t} dt$ (without the factor $1-e^{-\lambda c})$ and if you integrate by parts this gives the same answer as in Link 2.

The method used by Graham Kemp in Link2 is not applicable here since $X$ is a continuous random variable.

Kavi Rama Murthy
  • 311,013
2

I will give you a slightly easier way to think about the second method without indicators. Let's just split the integral domain. Kavis comment above is correct by the way.

\begin{align*} E(Y)&= \int_0^\infty P( Y > t) dt \\ &= 1) \int_0^c P( Y > t)dt + \int_c^{\infty}P(Y>t) dt \\ &= 2) \int_0^c P( X > t)dt + \int_c^{\infty}0 dt \\ &= 3) \int_0^c P(X > t)dt \\ &= 4) \int_0^c e^{-\lambda t}dt \end{align*}

Step $1)$ comes from simply splitting the integral over $(0,\infty)$ to $(0,c) \cup (c,\infty)$

Step $2)$ comes from thinking about what $Y$ is and how it can be in certain ranges. The only way it can be in between $0$ and $c$ is if it is $X$ taking this value. It also cannot be above $c$ because it is the minimum of something and $c$

Step $3)$ Just notes the integral of $0$ is $0$

Step $4)$ Just uses the CDF of exponential random variables.

OskarSzarowicz
  • 1,837