2 Poisson distributions time distribution given one of them occurs first
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Suppose that A and B are two independent Poisson distribution with parameter $lambda_a$ and $lambda_b$ denoting the number of, say the received emails per hour in two different computers. B is active after an hour later $t>1$. I want to find the expected value of the time that A receives the mail for the first time given A does not receive anything in the interval $t=[0,1]$ and before B does.
My attempt:
I use the same technique as the one Poisson rv as follows
The time distribution minus one is just($x =t-1$)
$P(X leq x) = $P("B does not receive anything but A receives at least one" ) = $(1-e^{-lambda_a x})e^{-lambda_b x} $
Based on this we can find the mean which is $frac{1}{lambda_b + lambda_a}- frac{1}{lambda_b} $ . But the correct answer is same as this except the second term takes positive sign. Is there any reasoning error here? If yes how to resolve this?
poisson-process exponential-distribution
asked Dec 9 '18 at 5:18
InterceptionInterception
233
$endgroup$
add a comment |
$begingroup$
Suppose that A and B are two independent Poisson distribution with parameter $lambda_a$ and $lambda_b$ denoting the number of, say the received emails per hour in two different computers. B is active after an hour later $t>1$. I want to find the expected value of the time that A receives the mail for the first time given A does not receive anything in the interval $t=[0,1]$ and before B does.
My attempt:
I use the same technique as the one Poisson rv as follows
The time distribution minus one is just($x =t-1$)
$P(X leq x) = $P("B does not receive anything but A receives at least one" ) = $(1-e^{-lambda_a x})e^{-lambda_b x} $
Based on this we can find the mean which is $frac{1}{lambda_b + lambda_a}- frac{1}{lambda_b} $ . But the correct answer is same as this except the second term takes positive sign. Is there any reasoning error here? If yes how to resolve this?
poisson-process exponential-distribution
asked Dec 9 '18 at 5:18
InterceptionInterception
233
$endgroup$
1
$begingroup$
It's quite difficult for me to parse your question. Could you please clarify it (rewrite parts of it)?
$endgroup$
– zoidberg
Dec 9 '18 at 5:28
add a comment |
$begingroup$
Suppose that A and B are two independent Poisson distribution with parameter $lambda_a$ and $lambda_b$ denoting the number of, say the received emails per hour in two different computers. B is active after an hour later $t>1$. I want to find the expected value of the time that A receives the mail for the first time given A does not receive anything in the interval $t=[0,1]$ and before B does.
My attempt:
I use the same technique as the one Poisson rv as follows
The time distribution minus one is just($x =t-1$)
$P(X leq x) = $P("B does not receive anything but A receives at least one" ) = $(1-e^{-lambda_a x})e^{-lambda_b x} $
Based on this we can find the mean which is $frac{1}{lambda_b + lambda_a}- frac{1}{lambda_b} $ . But the correct answer is same as this except the second term takes positive sign. Is there any reasoning error here? If yes how to resolve this?
poisson-process exponential-distribution
asked Dec 9 '18 at 5:18
InterceptionInterception
233
$endgroup$
Suppose that A and B are two independent Poisson distribution with parameter $lambda_a$ and $lambda_b$ denoting the number of, say the received emails per hour in two different computers. B is active after an hour later $t>1$. I want to find the expected value of the time that A receives the mail for the first time given A does not receive anything in the interval $t=[0,1]$ and before B does.
My attempt:
I use the same technique as the one Poisson rv as follows
The time distribution minus one is just($x =t-1$)
$P(X leq x) = $P("B does not receive anything but A receives at least one" ) = $(1-e^{-lambda_a x})e^{-lambda_b x} $
Based on this we can find the mean which is $frac{1}{lambda_b + lambda_a}- frac{1}{lambda_b} $ . But the correct answer is same as this except the second term takes positive sign. Is there any reasoning error here? If yes how to resolve this?
poisson-process exponential-distribution
poisson-process exponential-distribution
asked Dec 9 '18 at 5:18
InterceptionInterception
233
asked Dec 9 '18 at 5:18
InterceptionInterception
233
edited Dec 9 '18 at 6:15
Interception
asked Dec 9 '18 at 5:18
InterceptionInterception
233
asked Dec 9 '18 at 5:18
InterceptionInterception
233
asked Dec 9 '18 at 5:18
InterceptionInterception
233
233
1
$begingroup$
It's quite difficult for me to parse your question. Could you please clarify it (rewrite parts of it)?
$endgroup$
– zoidberg
Dec 9 '18 at 5:28
add a comment |
1
$begingroup$
It's quite difficult for me to parse your question. Could you please clarify it (rewrite parts of it)?
$endgroup$
– zoidberg
Dec 9 '18 at 5:28
1
1
$begingroup$
It's quite difficult for me to parse your question. Could you please clarify it (rewrite parts of it)?
$endgroup$
– zoidberg
Dec 9 '18 at 5:28
$begingroup$
It's quite difficult for me to parse your question. Could you please clarify it (rewrite parts of it)?
$endgroup$
– zoidberg
Dec 9 '18 at 5:28
add a comment |
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$begingroup$
It's quite difficult for me to parse your question. Could you please clarify it (rewrite parts of it)?
$endgroup$
– zoidberg
Dec 9 '18 at 5:28