Re: exponential R.V

On Oct 21, 4:32 pm, Jack Tomsky <jtom...@xxxxxxxxxxxxx> wrote:
The question is as follows

The durations of calls to a radio talk show are known
to be
exponentially distributed with a mean of 3 minutes.
Q1. Given that a call has already lasted 4 minutes,
what is the
probability that it will last at leat another 4
minutes?
Q2. Given that a call has already lasted 4 minutes,
what is the
expected remaining time until it ends?

So I'm going to solve these problems like this.

A1 to Q1) P[ X > 8 | X > 4]
A2 to Q2) E[ X | X > 4 }

Am I going to right way?

For an exponential, the conditional probability that X>t2, given that X>t1, is the unconditional probability that X>t2-t1. Thus,

Q1): exp(-4/3)

Q2): 4+E(X) = 4+3 = 7.

Jack


Actually, the answer to Q2) should be E[X | X > 4] - 4

= 4 + E[X] - 4 = E[X] = 3

since the question asks for the expected *remaining* time,
that is, the average time *beyond* the 4 minutes that the
call has already lasted.

--Dilip Sarwate

.

Relevant Pages

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    ... probability that it will last at leat another 4 ... For an exponential, the conditional probability ... the average time *beyond* the 4 minutes that ... --Dilip Sarwate ...
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  • Re: exponential R.V
    ... probability that it will last at leat another 4 ... the average time *beyond* the 4 minutes that the ... --Dilip Sarwate ... Can't one answer Q2 more directly by arguing that since the exponential ...
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  • Re: exponential R.V
    ... is the unconditional probability that X>t2-t1. ... that has the same distribution as X, and hence the same mean 3. ... Since the OP's Q2 asked for the average time till the call ends, ... one begins, as the OP proposed, by computing E, then ...
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