Find $mathbb{E}(W_{1}|min(W_{1},W_{2}))$ for $W_1$ and $W_2$ i.i.d. exponential
$begingroup$
Let $W_{1},W_{2}$ be independent, exponentially distributed random variables, $mathbb{E}(W_{i})=lambda$. Find $mathbb{E}(W_{1}|min(W_{1},W_{2})).$
I think I have the correct solution, but it's based solely on intuition.
Let $X=min(W_{1},W_{2})$. $$mathbb{E}(W_{1}|X)=mathbb{E}(W_{1}|X=W_{1},W_{1}<W_{2})mathbb{P}(W_{1}<W_{2})+mathbb{E}(W_{1}|X=W_{2},W_{2}<W_{1})mathbb{P}(W_{2}<W_{1})=frac{1}{2}(X+mathbb{E}(W_{1}|W_{1}>X))$$
And now the remaining expectation can be obtained by computing one integral and dividing it by the probability that the random variable $W_{1}$ is greater than $X$. Is this approach correct? Could someone explain, in mathematical terms, why the decomposition in the first line of the above reasoning is valid? (if it is).
probability-theory conditional-expectation
edited Jan 11 at 22:01
Did
248k23225463
asked Mar 11 '13 at 8:23
czachurczachur
1769
$endgroup$
add a comment |
$begingroup$
Let $W_{1},W_{2}$ be independent, exponentially distributed random variables, $mathbb{E}(W_{i})=lambda$. Find $mathbb{E}(W_{1}|min(W_{1},W_{2})).$
I think I have the correct solution, but it's based solely on intuition.
Let $X=min(W_{1},W_{2})$. $$mathbb{E}(W_{1}|X)=mathbb{E}(W_{1}|X=W_{1},W_{1}<W_{2})mathbb{P}(W_{1}<W_{2})+mathbb{E}(W_{1}|X=W_{2},W_{2}<W_{1})mathbb{P}(W_{2}<W_{1})=frac{1}{2}(X+mathbb{E}(W_{1}|W_{1}>X))$$
And now the remaining expectation can be obtained by computing one integral and dividing it by the probability that the random variable $W_{1}$ is greater than $X$. Is this approach correct? Could someone explain, in mathematical terms, why the decomposition in the first line of the above reasoning is valid? (if it is).
probability-theory conditional-expectation
edited Jan 11 at 22:01
Did
248k23225463
asked Mar 11 '13 at 8:23
czachurczachur
1769
$endgroup$
add a comment |
$begingroup$
Let $W_{1},W_{2}$ be independent, exponentially distributed random variables, $mathbb{E}(W_{i})=lambda$. Find $mathbb{E}(W_{1}|min(W_{1},W_{2})).$
I think I have the correct solution, but it's based solely on intuition.
Let $X=min(W_{1},W_{2})$. $$mathbb{E}(W_{1}|X)=mathbb{E}(W_{1}|X=W_{1},W_{1}<W_{2})mathbb{P}(W_{1}<W_{2})+mathbb{E}(W_{1}|X=W_{2},W_{2}<W_{1})mathbb{P}(W_{2}<W_{1})=frac{1}{2}(X+mathbb{E}(W_{1}|W_{1}>X))$$
And now the remaining expectation can be obtained by computing one integral and dividing it by the probability that the random variable $W_{1}$ is greater than $X$. Is this approach correct? Could someone explain, in mathematical terms, why the decomposition in the first line of the above reasoning is valid? (if it is).
probability-theory conditional-expectation
edited Jan 11 at 22:01
Did
248k23225463
asked Mar 11 '13 at 8:23
czachurczachur
1769
$endgroup$
Let $W_{1},W_{2}$ be independent, exponentially distributed random variables, $mathbb{E}(W_{i})=lambda$. Find $mathbb{E}(W_{1}|min(W_{1},W_{2})).$
I think I have the correct solution, but it's based solely on intuition.
Let $X=min(W_{1},W_{2})$. $$mathbb{E}(W_{1}|X)=mathbb{E}(W_{1}|X=W_{1},W_{1}<W_{2})mathbb{P}(W_{1}<W_{2})+mathbb{E}(W_{1}|X=W_{2},W_{2}<W_{1})mathbb{P}(W_{2}<W_{1})=frac{1}{2}(X+mathbb{E}(W_{1}|W_{1}>X))$$
And now the remaining expectation can be obtained by computing one integral and dividing it by the probability that the random variable $W_{1}$ is greater than $X$. Is this approach correct? Could someone explain, in mathematical terms, why the decomposition in the first line of the above reasoning is valid? (if it is).
probability-theory conditional-expectation
probability-theory conditional-expectation
edited Jan 11 at 22:01
Did
248k23225463
asked Mar 11 '13 at 8:23
czachurczachur
1769
edited Jan 11 at 22:01
Did
248k23225463
asked Mar 11 '13 at 8:23
czachurczachur
1769
edited Jan 11 at 22:01
Did
248k23225463
edited Jan 11 at 22:01
Did
248k23225463
edited Jan 11 at 22:01
Did
248k23225463
248k23225463
asked Mar 11 '13 at 8:23
czachurczachur
1769
asked Mar 11 '13 at 8:23
czachurczachur
1769
asked Mar 11 '13 at 8:23
czachurczachur
1769
1769
add a comment |
add a comment |
1 Answer
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$begingroup$
The loss of memory property of the exponential distribution implies that, conditionally on $X$, $W_1=X$ with probability $frac12$ and $W_1=X+W$ with probability $frac12$, where $W$ is independent of $X$ and exponential with parameter $lambda$. Hence,
$$
mathbb E(W_1mid X)=frac12X+frac12(X+mathbb E(W))=X+frac1{2lambda}.
$$
answered Mar 11 '13 at 8:38
DidDid
248k23225463
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1 Answer
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active
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1 Answer
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active
oldest
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$begingroup$
The loss of memory property of the exponential distribution implies that, conditionally on $X$, $W_1=X$ with probability $frac12$ and $W_1=X+W$ with probability $frac12$, where $W$ is independent of $X$ and exponential with parameter $lambda$. Hence,
$$
mathbb E(W_1mid X)=frac12X+frac12(X+mathbb E(W))=X+frac1{2lambda}.
$$
answered Mar 11 '13 at 8:38
DidDid
248k23225463
$endgroup$
add a comment |
$begingroup$
The loss of memory property of the exponential distribution implies that, conditionally on $X$, $W_1=X$ with probability $frac12$ and $W_1=X+W$ with probability $frac12$, where $W$ is independent of $X$ and exponential with parameter $lambda$. Hence,
$$
mathbb E(W_1mid X)=frac12X+frac12(X+mathbb E(W))=X+frac1{2lambda}.
$$
answered Mar 11 '13 at 8:38
DidDid
248k23225463
$endgroup$
add a comment |
$begingroup$
The loss of memory property of the exponential distribution implies that, conditionally on $X$, $W_1=X$ with probability $frac12$ and $W_1=X+W$ with probability $frac12$, where $W$ is independent of $X$ and exponential with parameter $lambda$. Hence,
$$
mathbb E(W_1mid X)=frac12X+frac12(X+mathbb E(W))=X+frac1{2lambda}.
$$
answered Mar 11 '13 at 8:38
DidDid
248k23225463
$endgroup$
The loss of memory property of the exponential distribution implies that, conditionally on $X$, $W_1=X$ with probability $frac12$ and $W_1=X+W$ with probability $frac12$, where $W$ is independent of $X$ and exponential with parameter $lambda$. Hence,
$$
mathbb E(W_1mid X)=frac12X+frac12(X+mathbb E(W))=X+frac1{2lambda}.
$$
answered Mar 11 '13 at 8:38
DidDid
248k23225463
answered Mar 11 '13 at 8:38
DidDid
248k23225463
answered Mar 11 '13 at 8:38
DidDid
248k23225463
answered Mar 11 '13 at 8:38
DidDid
248k23225463
248k23225463
add a comment |
add a comment |
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