2

Let $X$ be an exponential random variable with parameter $\lambda$. I have to find $\mathbb{E}[X\mid X<\alpha]$ where $\alpha >0$.

I must find $\mathbb{P}(X\mid X<\alpha)$ first.

What I did is as follows:

$\mathbb{P}(X\mid X<\alpha)=\dfrac{\mathbb{P}(X<t \;\text{and}\; X<\alpha)}{\mathbb{P}(X<\alpha)}$, where I assumed $t> \alpha$ then I get the conditional probability to be equal to $1$. I'm not sure if my assumption is right or wrong.

Michael Hardy
  • 1
Bhavish Suarez
  • 684
  • What do you mean by $\mathbb P(X\mid X < \alpha)$? Your calculation indicates the right side is a function of $t$ but the left side is not. Also, have you considered what happens if you assume that $0 < t < \alpha$? Your calculation for $t > \alpha$ is correct, but does not help in the least in solving the problem posed. – Dilip Sarwate May 24 '13 at 12:46
  • I was trying to find the conditional probability of $X$ given that it is less than $\alpha$, so that I could find the conditional expectation.

    Do you mean that I should consider both cases for this question? If I assume $t < \alpha$, I get the conditional probability to be $\dfrac{1-e^{-\lambda t}}{1-e^{-\lambda \alpha}}$.

     – Bhavish Suarez May 24 '13 at 13:02
  • There is no conditional probability of $X$; $X$ is a random variable and the only kinds of probabilities you should be talking of are probabilities of events such as ${X \in A}$, for example, $P{X < t}$ which happens to be $1-e^{-\lambda t}$ as you have correctly calculated. But this probability is not denoted $\mathbb P(X)$ which is meaningless. – Dilip Sarwate May 24 '13 at 13:25
  • Oh yes! I confused the notations...I actually found the conditional cumulative density function. Now I have to differentiate it. – Bhavish Suarez May 24 '13 at 13:52
  • 2
    @BhavishSuarez : There's no such thing as a "cumulative density function". The word "cumulative" contradicts the word "density". There is, however, a cumulative distribution function. Also, might you have meant $\mathbb P(X<t\mid X<\alpha)$? – Michael Hardy May 24 '13 at 14:28
  • 1
    If you want to evaluate the conditional expectation from basics, define rv $Y$ by $\Pr(Y\le 0)=0$, $\Pr(Y\le y)=1$ if $y\ge \alpha$, and $\Pr(Y\le y)=\Pr(X\le y|X\lt \alpha)$ if $0\lt y\lt \alpha$. You can compute this cdf, and then the expectation directly from the cdf. – André Nicolas May 24 '13 at 15:03
  • @MichaelHardy Yes, that's what I meant! – Bhavish Suarez May 24 '13 at 15:10
  • @AndréNicolas I did that and I got $\dfrac{1}{\lambda} - \dfrac{\alpha e^{-\lambda \alpha}}{1-e^{-\lambda \alpha}} $. However I'm not sure if it's right. – Bhavish Suarez May 24 '13 at 15:11
  • 1
    I get the same thing. Note also that it makes sense, should approach $\frac{1}{\lambda}$ as $\alpha$ gets large. – André Nicolas May 24 '13 at 15:30
  • I was confused about that for I thought that the answer should be $\alpha$ somehow. I don't why I thought so. Thanks a lot @AndréNicolas! – Bhavish Suarez May 24 '13 at 15:47

1 Answers1

2

The following is an approach that is more conceptual.

Let the conditional expectation of $X$ given that $X\lt \alpha$ be $A$.

Note that the expectation of $X$ given that $X\ge \alpha$ is $\alpha+\frac{1}{\lambda}$. This is a consequence of the fact that by the memorylessness property of the exponential, the additional "waiting time" given that we have waited an amount $\alpha$, is exponentially distributed with parameter $\lambda$.

It follows by the Law of Total Expectation that $$E(X)=\frac{1}{\lambda}=(1-e^{-\lambda\alpha})A+e^{-\lambda\alpha}\left(\alpha+\frac{1}{\lambda}\right).$$ Now we can solve for $A$.

Remark: Your approach from basics also works nicely. Define random variable $Y$ by $\Pr(Y\le y)=0$ if $y\le 0$, $\Pr(Y\le y)=1$ if $y\ge \alpha$, and $\Pr(Y\le y)=\Pr(X\le y|X\lt \alpha)$ otherwise. Then $$F_Y(y)=\frac{1-e^{-\lambda y}}{1-e^{-\lambda\alpha}}$$ for $0\lt y\lt \alpha$. Now we can compute $E(Y)$ in several ways. The most basic is to note that $f_Y(y)=\frac{\lambda e^{-\lambda y}}{1-e^{\lambda\alpha}}$ on $(0,\alpha)$ and $0$ elsewhere. Calculate $\int_0^\alpha y\frac{\lambda e^{-\lambda y}}{1-e^{\lambda\alpha}}\,dy$.

André Nicolas
  • 507,029