MHB Conditional exponential probability

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The discussion revolves around calculating the expected value of a random variable X, which follows different exponential distributions based on the occurrence of an event A. The law of total expectation is applied, leading to the formula E[X] = E[X|A] * P(A) + E[X|A^c] * P(A^c). Given that E[X|A] is 1/λ and E[X|A^c] is 1/μ, the expected value can be expressed as E[X] = (p/λ) + ((1 - p)/μ), where p is the probability of event A. Participants seek clarification on how to derive this expression using the parameters λ, μ, and the probability p. The conversation emphasizes understanding the application of conditional expectations in probability theory.
Longines
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Hello all,

I've been stuck on this question for a while and it's annoying the stew out of me!

I know it's a basic definition type of question, but I can't seem to understand it. Can any of you help?

Question:
Let X be a random variable and A be an event such that, conditional on A, X is exponential with parameter λ, and conditional on $A^c$ (A complement), X is exponential with parameter μ.
Write E[X] in terms of λ, μ and p, the probability of A
 
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Hello

Thankyou for sharing your problem with us at the MHB! :)

I suggest, you make a first step by writing down the law of total expectation:

$E[X] = \sum_{i=1}^{n} E[X|A_i]\cdot P(A_i)$

In this specific case the sum has only two terms ($n=2$)
 
lfdahl said:
Hello

Thankyou for sharing your problem with us at the MHB! :)

I suggest, you make a first step by writing down the law of total expectation:

$E[X] = \sum_{i=1}^{n} E[X|A_i]\cdot P(A_i)$

In this specific case the sum has only two terms ($n=2$)
I figured it had something to do with this, but I don't understand how I express it with lambda and such.

Could you please elaborate?
 
Longines said:
I figured it had something to do with this, but I don't understand how I express it with lambda and such.

Could you please elaborate?

The expectation of an exponential distribution with parameter $\lambda$ is $\frac 1 \lambda$.
So:
$$E[X|A] = \frac 1 \lambda$$
 
Longines said:
I figured it had something to do with this, but I don't understand how I express it with lambda and such.

Could you please elaborate?

First step:$E[X] = E[X|A] \cdot P(A)+E[X|A^c] \cdot P(A^c)$According to #4 you know $E[X|A]$ and $E[X|A^c]$ as $\frac{1}{\lambda}$ and $\frac{1}{\mu}$ respectively.If $P(A)=p$ then what is $P(A^c)$?

Now, try to express $E[X]$ in terms of $\lambda$, $\mu$ and $p$.
 
Longines said:
Hello all,

I've been stuck on this question for a while and it's annoying the hell out of me!

I know it's a basic definition type of question, but I can't seem to understand it. Can any of you help?

Question:
Let X be a random variable and A be an event such that, conditional on A, X is exponential with parameter λ, and conditional on $A^c$ (A complement), X is exponential with parameter μ.
Write E[X] in terms of λ, μ and p, the probability of A

If $P \{A\} = p$ and $P \{A^{c}\} = 1 - p$, then is...

$\displaystyle E \{X\} = \frac{p}{\lambda} + \frac{1 - p}{\mu}\ (1)$

Kind regards

$\chi$ $\sigma$
 
First trick I learned this one a long time ago and have used it to entertain and amuse young kids. Ask your friend to write down a three-digit number without showing it to you. Then ask him or her to rearrange the digits to form a new three-digit number. After that, write whichever is the larger number above the other number, and then subtract the smaller from the larger, making sure that you don't see any of the numbers. Then ask the young "victim" to tell you any two of the digits of the...

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