Exponential distribution of 2 independent events (expected value)
$begingroup$
X and Y are waiting time between phone calls for company A and B respectively and they are independent from each other. X and Y are exponentially distributed with expected waiting time of 10 min and 8 min respectively. What is the probability that the next phone call from either X or Y is within 5 min? What is the expected waiting time until the next phone call from either one of them?
I am not really sure what probability I am asked to look for. I'm guessing its P((X<5)U(Y<5)). For the expected waiting time I have no idea how to find it. Is it just E(X)+E(Y)?
probability exponential-distribution expected-value
asked Dec 8 '18 at 16:16
JunJun
31
$endgroup$
add a comment |
$begingroup$
X and Y are waiting time between phone calls for company A and B respectively and they are independent from each other. X and Y are exponentially distributed with expected waiting time of 10 min and 8 min respectively. What is the probability that the next phone call from either X or Y is within 5 min? What is the expected waiting time until the next phone call from either one of them?
I am not really sure what probability I am asked to look for. I'm guessing its P((X<5)U(Y<5)). For the expected waiting time I have no idea how to find it. Is it just E(X)+E(Y)?
probability exponential-distribution expected-value
asked Dec 8 '18 at 16:16
JunJun
31
$endgroup$
$begingroup$
Sorry but why do you think E(X)+E(Y) is even related to the question you are asked to solve?
$endgroup$
– Did
Dec 8 '18 at 16:20
$begingroup$
How else could I find the expected waiting time of both events happening? I think the union of the two events is not exponentially distributed right?
$endgroup$
– Jun
Dec 8 '18 at 16:26
$begingroup$
So... you have no idea about what the question is asking hence you answer... E(X)+E(Y)? Did I get you right?
$endgroup$
– Did
Dec 8 '18 at 16:29
$begingroup$
I think I understood the question but I got no clue how to find it. Yeah, E(X)+E(Y) was just a random stupid guess I came up with.
$endgroup$
– Jun
Dec 8 '18 at 16:31
add a comment |
$begingroup$
X and Y are waiting time between phone calls for company A and B respectively and they are independent from each other. X and Y are exponentially distributed with expected waiting time of 10 min and 8 min respectively. What is the probability that the next phone call from either X or Y is within 5 min? What is the expected waiting time until the next phone call from either one of them?
I am not really sure what probability I am asked to look for. I'm guessing its P((X<5)U(Y<5)). For the expected waiting time I have no idea how to find it. Is it just E(X)+E(Y)?
probability exponential-distribution expected-value
asked Dec 8 '18 at 16:16
JunJun
31
$endgroup$
X and Y are waiting time between phone calls for company A and B respectively and they are independent from each other. X and Y are exponentially distributed with expected waiting time of 10 min and 8 min respectively. What is the probability that the next phone call from either X or Y is within 5 min? What is the expected waiting time until the next phone call from either one of them?
I am not really sure what probability I am asked to look for. I'm guessing its P((X<5)U(Y<5)). For the expected waiting time I have no idea how to find it. Is it just E(X)+E(Y)?
probability exponential-distribution expected-value
probability exponential-distribution expected-value
asked Dec 8 '18 at 16:16
JunJun
31
asked Dec 8 '18 at 16:16
JunJun
31
asked Dec 8 '18 at 16:16
JunJun
31
asked Dec 8 '18 at 16:16
JunJun
31
asked Dec 8 '18 at 16:16
JunJun
31
31
$begingroup$
Sorry but why do you think E(X)+E(Y) is even related to the question you are asked to solve?
$endgroup$
– Did
Dec 8 '18 at 16:20
$begingroup$
How else could I find the expected waiting time of both events happening? I think the union of the two events is not exponentially distributed right?
$endgroup$
– Jun
Dec 8 '18 at 16:26
$begingroup$
So... you have no idea about what the question is asking hence you answer... E(X)+E(Y)? Did I get you right?
$endgroup$
– Did
Dec 8 '18 at 16:29
$begingroup$
I think I understood the question but I got no clue how to find it. Yeah, E(X)+E(Y) was just a random stupid guess I came up with.
$endgroup$
– Jun
Dec 8 '18 at 16:31
add a comment |
$begingroup$
Sorry but why do you think E(X)+E(Y) is even related to the question you are asked to solve?
$endgroup$
– Did
Dec 8 '18 at 16:20
$begingroup$
How else could I find the expected waiting time of both events happening? I think the union of the two events is not exponentially distributed right?
$endgroup$
– Jun
Dec 8 '18 at 16:26
$begingroup$
So... you have no idea about what the question is asking hence you answer... E(X)+E(Y)? Did I get you right?
$endgroup$
– Did
Dec 8 '18 at 16:29
$begingroup$
I think I understood the question but I got no clue how to find it. Yeah, E(X)+E(Y) was just a random stupid guess I came up with.
$endgroup$
– Jun
Dec 8 '18 at 16:31
$begingroup$
Sorry but why do you think E(X)+E(Y) is even related to the question you are asked to solve?
$endgroup$
– Did
Dec 8 '18 at 16:20
$begingroup$
Sorry but why do you think E(X)+E(Y) is even related to the question you are asked to solve?
$endgroup$
– Did
Dec 8 '18 at 16:20
$begingroup$
How else could I find the expected waiting time of both events happening? I think the union of the two events is not exponentially distributed right?
$endgroup$
– Jun
Dec 8 '18 at 16:26
$begingroup$
How else could I find the expected waiting time of both events happening? I think the union of the two events is not exponentially distributed right?
$endgroup$
– Jun
Dec 8 '18 at 16:26
$begingroup$
So... you have no idea about what the question is asking hence you answer... E(X)+E(Y)? Did I get you right?
$endgroup$
– Did
Dec 8 '18 at 16:29
$begingroup$
So... you have no idea about what the question is asking hence you answer... E(X)+E(Y)? Did I get you right?
$endgroup$
– Did
Dec 8 '18 at 16:29
$begingroup$
I think I understood the question but I got no clue how to find it. Yeah, E(X)+E(Y) was just a random stupid guess I came up with.
$endgroup$
– Jun
Dec 8 '18 at 16:31
$begingroup$
I think I understood the question but I got no clue how to find it. Yeah, E(X)+E(Y) was just a random stupid guess I came up with.
$endgroup$
– Jun
Dec 8 '18 at 16:31
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The next phone call is the minimum of $X$ and $Y$; write $Z=Xwedge Y$. The intuition behind this should be fairly obvious. Now, the distribution of the minimum of two independent exponentially distributed random variables is again exponential, with rate the sum of rates of the two. To see this, suppose $Xsimmathsf{Exp}(lambda)$ and $Ysimmathsf{Exp}(mu)$, then for any $t>0$ we have
begin{align}
mathbb P(Z>t) &= mathbb P(X>t, Y>t)\
&= mathbb P(X>t)mathbb P(Y>t)\
&= e^{-lambda t}e^{-mu t}\
&= e^{-(lambda+mu)t},
end{align}
so that $Zsimmathsf{Exp}(lambda+mu)$. Plugging in $lambda = frac1{10}$, $mu=frac18$, and $t=5$, we have
$$
mathbb P(Zleqslant 5) = 1 - e^{-left(frac1{10}+frac18 right)cdot5} = 1-e^{-frac98}.
$$
The expected value of $Z$ is simply $$left(frac1{10}+frac18 right)^{-1}=frac{40}9. $$
answered Dec 8 '18 at 22:47
Math1000Math1000
19.3k31745
$endgroup$
$begingroup$
But for the next phone call, between X and Y, only one of them needs to be within 5 minutes to fulfill the condition. Shouldn't it be Z= X ∨ Y ?
$endgroup$
– Jun
Dec 9 '18 at 0:49
$begingroup$
Only one of them - which would be the minimum. The maximum would be for both of them.
$endgroup$
– Math1000
Dec 9 '18 at 3:25
add a comment |
$begingroup$
I think you are searching for the right probability. And yes, in that case you can exploit the linearity of the expected value just as you wrote.
However, @Did is right that the expected value is not related to your problem. Your problem might be solved by simply trying to calculate the stated probability and using $P((X < 5) lor (Y<5)) = P(X<5) + P(Y<5) - P((X<5)land(Y<5))$
answered Dec 8 '18 at 16:19
Thomas LangThomas Lang
1624
$endgroup$
$begingroup$
It's just a random crazy guess for the expected value because I couldn't think of other ways to look for expected value of 2 independent events from what I've been taught.
$endgroup$
– Jun
Dec 8 '18 at 16:29
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The next phone call is the minimum of $X$ and $Y$; write $Z=Xwedge Y$. The intuition behind this should be fairly obvious. Now, the distribution of the minimum of two independent exponentially distributed random variables is again exponential, with rate the sum of rates of the two. To see this, suppose $Xsimmathsf{Exp}(lambda)$ and $Ysimmathsf{Exp}(mu)$, then for any $t>0$ we have
begin{align}
mathbb P(Z>t) &= mathbb P(X>t, Y>t)\
&= mathbb P(X>t)mathbb P(Y>t)\
&= e^{-lambda t}e^{-mu t}\
&= e^{-(lambda+mu)t},
end{align}
so that $Zsimmathsf{Exp}(lambda+mu)$. Plugging in $lambda = frac1{10}$, $mu=frac18$, and $t=5$, we have
$$
mathbb P(Zleqslant 5) = 1 - e^{-left(frac1{10}+frac18 right)cdot5} = 1-e^{-frac98}.
$$
The expected value of $Z$ is simply $$left(frac1{10}+frac18 right)^{-1}=frac{40}9. $$
answered Dec 8 '18 at 22:47
Math1000Math1000
19.3k31745
$endgroup$
$begingroup$
But for the next phone call, between X and Y, only one of them needs to be within 5 minutes to fulfill the condition. Shouldn't it be Z= X ∨ Y ?
$endgroup$
– Jun
Dec 9 '18 at 0:49
$begingroup$
Only one of them - which would be the minimum. The maximum would be for both of them.
$endgroup$
– Math1000
Dec 9 '18 at 3:25
add a comment |
$begingroup$
The next phone call is the minimum of $X$ and $Y$; write $Z=Xwedge Y$. The intuition behind this should be fairly obvious. Now, the distribution of the minimum of two independent exponentially distributed random variables is again exponential, with rate the sum of rates of the two. To see this, suppose $Xsimmathsf{Exp}(lambda)$ and $Ysimmathsf{Exp}(mu)$, then for any $t>0$ we have
begin{align}
mathbb P(Z>t) &= mathbb P(X>t, Y>t)\
&= mathbb P(X>t)mathbb P(Y>t)\
&= e^{-lambda t}e^{-mu t}\
&= e^{-(lambda+mu)t},
end{align}
so that $Zsimmathsf{Exp}(lambda+mu)$. Plugging in $lambda = frac1{10}$, $mu=frac18$, and $t=5$, we have
$$
mathbb P(Zleqslant 5) = 1 - e^{-left(frac1{10}+frac18 right)cdot5} = 1-e^{-frac98}.
$$
The expected value of $Z$ is simply $$left(frac1{10}+frac18 right)^{-1}=frac{40}9. $$
answered Dec 8 '18 at 22:47
Math1000Math1000
19.3k31745
$endgroup$
$begingroup$
But for the next phone call, between X and Y, only one of them needs to be within 5 minutes to fulfill the condition. Shouldn't it be Z= X ∨ Y ?
$endgroup$
– Jun
Dec 9 '18 at 0:49
$begingroup$
Only one of them - which would be the minimum. The maximum would be for both of them.
$endgroup$
– Math1000
Dec 9 '18 at 3:25
add a comment |
$begingroup$
The next phone call is the minimum of $X$ and $Y$; write $Z=Xwedge Y$. The intuition behind this should be fairly obvious. Now, the distribution of the minimum of two independent exponentially distributed random variables is again exponential, with rate the sum of rates of the two. To see this, suppose $Xsimmathsf{Exp}(lambda)$ and $Ysimmathsf{Exp}(mu)$, then for any $t>0$ we have
begin{align}
mathbb P(Z>t) &= mathbb P(X>t, Y>t)\
&= mathbb P(X>t)mathbb P(Y>t)\
&= e^{-lambda t}e^{-mu t}\
&= e^{-(lambda+mu)t},
end{align}
so that $Zsimmathsf{Exp}(lambda+mu)$. Plugging in $lambda = frac1{10}$, $mu=frac18$, and $t=5$, we have
$$
mathbb P(Zleqslant 5) = 1 - e^{-left(frac1{10}+frac18 right)cdot5} = 1-e^{-frac98}.
$$
The expected value of $Z$ is simply $$left(frac1{10}+frac18 right)^{-1}=frac{40}9. $$
answered Dec 8 '18 at 22:47
Math1000Math1000
19.3k31745
$endgroup$
The next phone call is the minimum of $X$ and $Y$; write $Z=Xwedge Y$. The intuition behind this should be fairly obvious. Now, the distribution of the minimum of two independent exponentially distributed random variables is again exponential, with rate the sum of rates of the two. To see this, suppose $Xsimmathsf{Exp}(lambda)$ and $Ysimmathsf{Exp}(mu)$, then for any $t>0$ we have
begin{align}
mathbb P(Z>t) &= mathbb P(X>t, Y>t)\
&= mathbb P(X>t)mathbb P(Y>t)\
&= e^{-lambda t}e^{-mu t}\
&= e^{-(lambda+mu)t},
end{align}
so that $Zsimmathsf{Exp}(lambda+mu)$. Plugging in $lambda = frac1{10}$, $mu=frac18$, and $t=5$, we have
$$
mathbb P(Zleqslant 5) = 1 - e^{-left(frac1{10}+frac18 right)cdot5} = 1-e^{-frac98}.
$$
The expected value of $Z$ is simply $$left(frac1{10}+frac18 right)^{-1}=frac{40}9. $$
answered Dec 8 '18 at 22:47
Math1000Math1000
19.3k31745
answered Dec 8 '18 at 22:47
Math1000Math1000
19.3k31745
answered Dec 8 '18 at 22:47
Math1000Math1000
19.3k31745
answered Dec 8 '18 at 22:47
Math1000Math1000
19.3k31745
19.3k31745
$begingroup$
But for the next phone call, between X and Y, only one of them needs to be within 5 minutes to fulfill the condition. Shouldn't it be Z= X ∨ Y ?
$endgroup$
– Jun
Dec 9 '18 at 0:49
$begingroup$
Only one of them - which would be the minimum. The maximum would be for both of them.
$endgroup$
– Math1000
Dec 9 '18 at 3:25
add a comment |
$begingroup$
But for the next phone call, between X and Y, only one of them needs to be within 5 minutes to fulfill the condition. Shouldn't it be Z= X ∨ Y ?
$endgroup$
– Jun
Dec 9 '18 at 0:49
$begingroup$
Only one of them - which would be the minimum. The maximum would be for both of them.
$endgroup$
– Math1000
Dec 9 '18 at 3:25
$begingroup$
But for the next phone call, between X and Y, only one of them needs to be within 5 minutes to fulfill the condition. Shouldn't it be Z= X ∨ Y ?
$endgroup$
– Jun
Dec 9 '18 at 0:49
$begingroup$
But for the next phone call, between X and Y, only one of them needs to be within 5 minutes to fulfill the condition. Shouldn't it be Z= X ∨ Y ?
$endgroup$
– Jun
Dec 9 '18 at 0:49
$begingroup$
Only one of them - which would be the minimum. The maximum would be for both of them.
$endgroup$
– Math1000
Dec 9 '18 at 3:25
$begingroup$
Only one of them - which would be the minimum. The maximum would be for both of them.
$endgroup$
– Math1000
Dec 9 '18 at 3:25
add a comment |
$begingroup$
I think you are searching for the right probability. And yes, in that case you can exploit the linearity of the expected value just as you wrote.
However, @Did is right that the expected value is not related to your problem. Your problem might be solved by simply trying to calculate the stated probability and using $P((X < 5) lor (Y<5)) = P(X<5) + P(Y<5) - P((X<5)land(Y<5))$
answered Dec 8 '18 at 16:19
Thomas LangThomas Lang
1624
$endgroup$
$begingroup$
It's just a random crazy guess for the expected value because I couldn't think of other ways to look for expected value of 2 independent events from what I've been taught.
$endgroup$
– Jun
Dec 8 '18 at 16:29
add a comment |
$begingroup$
I think you are searching for the right probability. And yes, in that case you can exploit the linearity of the expected value just as you wrote.
However, @Did is right that the expected value is not related to your problem. Your problem might be solved by simply trying to calculate the stated probability and using $P((X < 5) lor (Y<5)) = P(X<5) + P(Y<5) - P((X<5)land(Y<5))$
answered Dec 8 '18 at 16:19
Thomas LangThomas Lang
1624
$endgroup$
$begingroup$
It's just a random crazy guess for the expected value because I couldn't think of other ways to look for expected value of 2 independent events from what I've been taught.
$endgroup$
– Jun
Dec 8 '18 at 16:29
add a comment |
$begingroup$
I think you are searching for the right probability. And yes, in that case you can exploit the linearity of the expected value just as you wrote.
However, @Did is right that the expected value is not related to your problem. Your problem might be solved by simply trying to calculate the stated probability and using $P((X < 5) lor (Y<5)) = P(X<5) + P(Y<5) - P((X<5)land(Y<5))$
answered Dec 8 '18 at 16:19
Thomas LangThomas Lang
1624
$endgroup$
I think you are searching for the right probability. And yes, in that case you can exploit the linearity of the expected value just as you wrote.
However, @Did is right that the expected value is not related to your problem. Your problem might be solved by simply trying to calculate the stated probability and using $P((X < 5) lor (Y<5)) = P(X<5) + P(Y<5) - P((X<5)land(Y<5))$
answered Dec 8 '18 at 16:19
Thomas LangThomas Lang
1624
edited Dec 8 '18 at 16:25
answered Dec 8 '18 at 16:19
Thomas LangThomas Lang
1624
answered Dec 8 '18 at 16:19
Thomas LangThomas Lang
1624
answered Dec 8 '18 at 16:19
Thomas LangThomas Lang
1624
1624
$begingroup$
It's just a random crazy guess for the expected value because I couldn't think of other ways to look for expected value of 2 independent events from what I've been taught.
$endgroup$
– Jun
Dec 8 '18 at 16:29
add a comment |
$begingroup$
It's just a random crazy guess for the expected value because I couldn't think of other ways to look for expected value of 2 independent events from what I've been taught.
$endgroup$
– Jun
Dec 8 '18 at 16:29
$begingroup$
It's just a random crazy guess for the expected value because I couldn't think of other ways to look for expected value of 2 independent events from what I've been taught.
$endgroup$
– Jun
Dec 8 '18 at 16:29
$begingroup$
It's just a random crazy guess for the expected value because I couldn't think of other ways to look for expected value of 2 independent events from what I've been taught.
$endgroup$
– Jun
Dec 8 '18 at 16:29
add a comment |
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$begingroup$
Sorry but why do you think E(X)+E(Y) is even related to the question you are asked to solve?
$endgroup$
– Did
Dec 8 '18 at 16:20
$begingroup$
How else could I find the expected waiting time of both events happening? I think the union of the two events is not exponentially distributed right?
$endgroup$
– Jun
Dec 8 '18 at 16:26
$begingroup$
So... you have no idea about what the question is asking hence you answer... E(X)+E(Y)? Did I get you right?
$endgroup$
– Did
Dec 8 '18 at 16:29
$begingroup$
I think I understood the question but I got no clue how to find it. Yeah, E(X)+E(Y) was just a random stupid guess I came up with.
$endgroup$
– Jun
Dec 8 '18 at 16:31