Compute $P{tlt Xlt Y}$ for $t ge 0$, when $X$ and $Y$ are independent exponential random variables, possibly...
$begingroup$
What is $P(t lt X lt Y)$ for $tgt 0$, when X and Y are independent exponential variables with the parameters $lambda$ and $mu$ respectively?
Since X and Y are independent, I got joint PDF to be $lambda cdotmucdot
e^{-lambdacdot x-mucdot y}$.
Now that I'm looking for $P(t lt X lt Y)$, using the joint PDF, I set it as:
$$
P(t<X<Y)=int_0^inftyint_t^y lambda cdotmucdot
e^{-lambdacdot x-mucdot y}dxdy
$$
Is this correct? I'm trying to solve it using the online integral calculator but it's not getting the answer.
Edit 1: Here are the changes and result I came up with
Edit 2: Fixed my calculation error and got a confident answer:
$$
P(t<X<Y)=int_t^inftyint_t^y lambda cdotmucdot
e^{-lambdacdot x-mucdot y}dxdy=frac{lambda}{lambda+mu}e^{-t(lambda+mu)}
$$
I would really appreciate if someone could check if this is correct. Thank you!
probability-theory probability-distributions
asked Jan 11 at 15:36
hilageryolohilageryolo
173
$endgroup$
add a comment |
$begingroup$
What is $P(t lt X lt Y)$ for $tgt 0$, when X and Y are independent exponential variables with the parameters $lambda$ and $mu$ respectively?
Since X and Y are independent, I got joint PDF to be $lambda cdotmucdot
e^{-lambdacdot x-mucdot y}$.
Now that I'm looking for $P(t lt X lt Y)$, using the joint PDF, I set it as:
$$
P(t<X<Y)=int_0^inftyint_t^y lambda cdotmucdot
e^{-lambdacdot x-mucdot y}dxdy
$$
Is this correct? I'm trying to solve it using the online integral calculator but it's not getting the answer.
Edit 1: Here are the changes and result I came up with
Edit 2: Fixed my calculation error and got a confident answer:
$$
P(t<X<Y)=int_t^inftyint_t^y lambda cdotmucdot
e^{-lambdacdot x-mucdot y}dxdy=frac{lambda}{lambda+mu}e^{-t(lambda+mu)}
$$
I would really appreciate if someone could check if this is correct. Thank you!
probability-theory probability-distributions
asked Jan 11 at 15:36
hilageryolohilageryolo
173
$endgroup$
1
$begingroup$
Well, even without the help of online tools, you should be able to compute $$int_t^ylambda e^{-lambda x}dx$$ and to plug the result in the integral with respect to $x$...
$endgroup$
– Did
Jan 11 at 16:11
$begingroup$
Observe that $t < X < Y$ implies $t < Y$ and hence, the "domain of integration" (as Americans call it) is the set of $y$ such that $t < y.$ Having such a fix $y,$ the domain of integration of $X$ is simply $t < x < y.$ What I am expressing here is that your double integral is wrong.
$endgroup$
– Will M.
Jan 11 at 16:51
$begingroup$
@WillM. Since $t<y$, would the limit of the $dy$ integral be from t to $infty$?
$endgroup$
– hilageryolo
Jan 11 at 20:03
add a comment |
$begingroup$
What is $P(t lt X lt Y)$ for $tgt 0$, when X and Y are independent exponential variables with the parameters $lambda$ and $mu$ respectively?
Since X and Y are independent, I got joint PDF to be $lambda cdotmucdot
e^{-lambdacdot x-mucdot y}$.
Now that I'm looking for $P(t lt X lt Y)$, using the joint PDF, I set it as:
$$
P(t<X<Y)=int_0^inftyint_t^y lambda cdotmucdot
e^{-lambdacdot x-mucdot y}dxdy
$$
Is this correct? I'm trying to solve it using the online integral calculator but it's not getting the answer.
Edit 1: Here are the changes and result I came up with
Edit 2: Fixed my calculation error and got a confident answer:
$$
P(t<X<Y)=int_t^inftyint_t^y lambda cdotmucdot
e^{-lambdacdot x-mucdot y}dxdy=frac{lambda}{lambda+mu}e^{-t(lambda+mu)}
$$
I would really appreciate if someone could check if this is correct. Thank you!
probability-theory probability-distributions
asked Jan 11 at 15:36
hilageryolohilageryolo
173
$endgroup$
What is $P(t lt X lt Y)$ for $tgt 0$, when X and Y are independent exponential variables with the parameters $lambda$ and $mu$ respectively?
Since X and Y are independent, I got joint PDF to be $lambda cdotmucdot
e^{-lambdacdot x-mucdot y}$.
Now that I'm looking for $P(t lt X lt Y)$, using the joint PDF, I set it as:
$$
P(t<X<Y)=int_0^inftyint_t^y lambda cdotmucdot
e^{-lambdacdot x-mucdot y}dxdy
$$
Is this correct? I'm trying to solve it using the online integral calculator but it's not getting the answer.
Edit 1: Here are the changes and result I came up with
Edit 2: Fixed my calculation error and got a confident answer:
$$
P(t<X<Y)=int_t^inftyint_t^y lambda cdotmucdot
e^{-lambdacdot x-mucdot y}dxdy=frac{lambda}{lambda+mu}e^{-t(lambda+mu)}
$$
I would really appreciate if someone could check if this is correct. Thank you!
probability-theory probability-distributions
probability-theory probability-distributions
asked Jan 11 at 15:36
hilageryolohilageryolo
173
asked Jan 11 at 15:36
hilageryolohilageryolo
173
edited Jan 11 at 21:58
hilageryolo
asked Jan 11 at 15:36
hilageryolohilageryolo
173
asked Jan 11 at 15:36
hilageryolohilageryolo
173
asked Jan 11 at 15:36
hilageryolohilageryolo
173
173
1
$begingroup$
Well, even without the help of online tools, you should be able to compute $$int_t^ylambda e^{-lambda x}dx$$ and to plug the result in the integral with respect to $x$...
$endgroup$
– Did
Jan 11 at 16:11
$begingroup$
Observe that $t < X < Y$ implies $t < Y$ and hence, the "domain of integration" (as Americans call it) is the set of $y$ such that $t < y.$ Having such a fix $y,$ the domain of integration of $X$ is simply $t < x < y.$ What I am expressing here is that your double integral is wrong.
$endgroup$
– Will M.
Jan 11 at 16:51
$begingroup$
@WillM. Since $t<y$, would the limit of the $dy$ integral be from t to $infty$?
$endgroup$
– hilageryolo
Jan 11 at 20:03
add a comment |
1
$begingroup$
Well, even without the help of online tools, you should be able to compute $$int_t^ylambda e^{-lambda x}dx$$ and to plug the result in the integral with respect to $x$...
$endgroup$
– Did
Jan 11 at 16:11
$begingroup$
Observe that $t < X < Y$ implies $t < Y$ and hence, the "domain of integration" (as Americans call it) is the set of $y$ such that $t < y.$ Having such a fix $y,$ the domain of integration of $X$ is simply $t < x < y.$ What I am expressing here is that your double integral is wrong.
$endgroup$
– Will M.
Jan 11 at 16:51
$begingroup$
@WillM. Since $t<y$, would the limit of the $dy$ integral be from t to $infty$?
$endgroup$
– hilageryolo
Jan 11 at 20:03
1
1
$begingroup$
Well, even without the help of online tools, you should be able to compute $$int_t^ylambda e^{-lambda x}dx$$ and to plug the result in the integral with respect to $x$...
$endgroup$
– Did
Jan 11 at 16:11
$begingroup$
Well, even without the help of online tools, you should be able to compute $$int_t^ylambda e^{-lambda x}dx$$ and to plug the result in the integral with respect to $x$...
$endgroup$
– Did
Jan 11 at 16:11
$begingroup$
Observe that $t < X < Y$ implies $t < Y$ and hence, the "domain of integration" (as Americans call it) is the set of $y$ such that $t < y.$ Having such a fix $y,$ the domain of integration of $X$ is simply $t < x < y.$ What I am expressing here is that your double integral is wrong.
$endgroup$
– Will M.
Jan 11 at 16:51
$begingroup$
Observe that $t < X < Y$ implies $t < Y$ and hence, the "domain of integration" (as Americans call it) is the set of $y$ such that $t < y.$ Having such a fix $y,$ the domain of integration of $X$ is simply $t < x < y.$ What I am expressing here is that your double integral is wrong.
$endgroup$
– Will M.
Jan 11 at 16:51
$begingroup$
@WillM. Since $t<y$, would the limit of the $dy$ integral be from t to $infty$?
$endgroup$
– hilageryolo
Jan 11 at 20:03
$begingroup$
@WillM. Since $t<y$, would the limit of the $dy$ integral be from t to $infty$?
$endgroup$
– hilageryolo
Jan 11 at 20:03
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Try this in Mathematica
Probability[ t < x < y, {x [Distributed] ExponentialDistribution[[Lambda]], y [Distributed] ExponentialDistribution[[Mu]]}]
You will get
$$
begin{array}{cc}
& left{
begin{array}{cc}
frac{lambda }{lambda +mu } & t<0 \
frac{lambda e^{-t (lambda +mu )}}{lambda +mu } & t ge 0 \
end{array}
right.
\
end{array}
$$
answered Jan 11 at 21:17
ablmfablmf
2,56842452
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Try this in Mathematica
Probability[ t < x < y, {x [Distributed] ExponentialDistribution[[Lambda]], y [Distributed] ExponentialDistribution[[Mu]]}]
You will get
$$
begin{array}{cc}
& left{
begin{array}{cc}
frac{lambda }{lambda +mu } & t<0 \
frac{lambda e^{-t (lambda +mu )}}{lambda +mu } & t ge 0 \
end{array}
right.
\
end{array}
$$
answered Jan 11 at 21:17
ablmfablmf
2,56842452
$endgroup$
add a comment |
$begingroup$
Try this in Mathematica
Probability[ t < x < y, {x [Distributed] ExponentialDistribution[[Lambda]], y [Distributed] ExponentialDistribution[[Mu]]}]
You will get
$$
begin{array}{cc}
& left{
begin{array}{cc}
frac{lambda }{lambda +mu } & t<0 \
frac{lambda e^{-t (lambda +mu )}}{lambda +mu } & t ge 0 \
end{array}
right.
\
end{array}
$$
answered Jan 11 at 21:17
ablmfablmf
2,56842452
$endgroup$
add a comment |
$begingroup$
Try this in Mathematica
Probability[ t < x < y, {x [Distributed] ExponentialDistribution[[Lambda]], y [Distributed] ExponentialDistribution[[Mu]]}]
You will get
$$
begin{array}{cc}
& left{
begin{array}{cc}
frac{lambda }{lambda +mu } & t<0 \
frac{lambda e^{-t (lambda +mu )}}{lambda +mu } & t ge 0 \
end{array}
right.
\
end{array}
$$
answered Jan 11 at 21:17
ablmfablmf
2,56842452
$endgroup$
Try this in Mathematica
Probability[ t < x < y, {x [Distributed] ExponentialDistribution[[Lambda]], y [Distributed] ExponentialDistribution[[Mu]]}]
You will get
$$
begin{array}{cc}
& left{
begin{array}{cc}
frac{lambda }{lambda +mu } & t<0 \
frac{lambda e^{-t (lambda +mu )}}{lambda +mu } & t ge 0 \
end{array}
right.
\
end{array}
$$
answered Jan 11 at 21:17
ablmfablmf
2,56842452
answered Jan 11 at 21:17
ablmfablmf
2,56842452
answered Jan 11 at 21:17
ablmfablmf
2,56842452
answered Jan 11 at 21:17
ablmfablmf
2,56842452
2,56842452
add a comment |
add a comment |
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1
$begingroup$
Well, even without the help of online tools, you should be able to compute $$int_t^ylambda e^{-lambda x}dx$$ and to plug the result in the integral with respect to $x$...
$endgroup$
– Did
Jan 11 at 16:11
$begingroup$
Observe that $t < X < Y$ implies $t < Y$ and hence, the "domain of integration" (as Americans call it) is the set of $y$ such that $t < y.$ Having such a fix $y,$ the domain of integration of $X$ is simply $t < x < y.$ What I am expressing here is that your double integral is wrong.
$endgroup$
– Will M.
Jan 11 at 16:51
$begingroup$
@WillM. Since $t<y$, would the limit of the $dy$ integral be from t to $infty$?
$endgroup$
– hilageryolo
Jan 11 at 20:03