probability re: comparing 2 iid exponential random variables
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Let $X, Y$ be iid exponential random variables with parameter $1$. Then, what is the probability that $X < Y + 1$?
I know how to compute $P(X < Y)$ (from integrating the joint PDF, which is the same as the product of the marginal PDFs for X and Y because X and Y are independent, and the marginal PDF is the PDF of an exponential random variable with parameter $1$).
Maybe I'm missing something obvious, but the $+ 1$ is really tripping me up.
probability probability-distributions exponential-distribution
asked Dec 9 '18 at 17:18
0k330k33
12010
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add a comment |
$begingroup$
Let $X, Y$ be iid exponential random variables with parameter $1$. Then, what is the probability that $X < Y + 1$?
I know how to compute $P(X < Y)$ (from integrating the joint PDF, which is the same as the product of the marginal PDFs for X and Y because X and Y are independent, and the marginal PDF is the PDF of an exponential random variable with parameter $1$).
Maybe I'm missing something obvious, but the $+ 1$ is really tripping me up.
probability probability-distributions exponential-distribution
asked Dec 9 '18 at 17:18
0k330k33
12010
$endgroup$
add a comment |
$begingroup$
Let $X, Y$ be iid exponential random variables with parameter $1$. Then, what is the probability that $X < Y + 1$?
I know how to compute $P(X < Y)$ (from integrating the joint PDF, which is the same as the product of the marginal PDFs for X and Y because X and Y are independent, and the marginal PDF is the PDF of an exponential random variable with parameter $1$).
Maybe I'm missing something obvious, but the $+ 1$ is really tripping me up.
probability probability-distributions exponential-distribution
asked Dec 9 '18 at 17:18
0k330k33
12010
$endgroup$
Let $X, Y$ be iid exponential random variables with parameter $1$. Then, what is the probability that $X < Y + 1$?
I know how to compute $P(X < Y)$ (from integrating the joint PDF, which is the same as the product of the marginal PDFs for X and Y because X and Y are independent, and the marginal PDF is the PDF of an exponential random variable with parameter $1$).
Maybe I'm missing something obvious, but the $+ 1$ is really tripping me up.
probability probability-distributions exponential-distribution
probability probability-distributions exponential-distribution
asked Dec 9 '18 at 17:18
0k330k33
12010
asked Dec 9 '18 at 17:18
0k330k33
12010
asked Dec 9 '18 at 17:18
0k330k33
12010
asked Dec 9 '18 at 17:18
0k330k33
12010
asked Dec 9 '18 at 17:18
0k330k33
12010
12010
add a comment |
add a comment |
1 Answer
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Given that you know how to integrate the joint density, then all you have to do is modify the region of integration suitably: The inequality $X < Y + 1$ with the added conditions $X > 0$, $Y > 0$, would represent what region of the $(X,Y)$ coordinate plane? Plot the equation $Y = X - 1$. What does that look like? Now, sketch the region satisfying the inequalities. Then consider how to set up an iterated integral over which you would integrate the joint density: one order of integration will be more inconvenient than the other, so choose carefully.
answered Dec 9 '18 at 17:27
heropupheropup
63k66199
$endgroup$
$begingroup$
Thank you! For some reason I thought the distribution itself would change, not just the integration limits.
$endgroup$
– 0k33
Dec 9 '18 at 19:09
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Given that you know how to integrate the joint density, then all you have to do is modify the region of integration suitably: The inequality $X < Y + 1$ with the added conditions $X > 0$, $Y > 0$, would represent what region of the $(X,Y)$ coordinate plane? Plot the equation $Y = X - 1$. What does that look like? Now, sketch the region satisfying the inequalities. Then consider how to set up an iterated integral over which you would integrate the joint density: one order of integration will be more inconvenient than the other, so choose carefully.
answered Dec 9 '18 at 17:27
heropupheropup
63k66199
$endgroup$
$begingroup$
Thank you! For some reason I thought the distribution itself would change, not just the integration limits.
$endgroup$
– 0k33
Dec 9 '18 at 19:09
add a comment |
$begingroup$
Given that you know how to integrate the joint density, then all you have to do is modify the region of integration suitably: The inequality $X < Y + 1$ with the added conditions $X > 0$, $Y > 0$, would represent what region of the $(X,Y)$ coordinate plane? Plot the equation $Y = X - 1$. What does that look like? Now, sketch the region satisfying the inequalities. Then consider how to set up an iterated integral over which you would integrate the joint density: one order of integration will be more inconvenient than the other, so choose carefully.
answered Dec 9 '18 at 17:27
heropupheropup
63k66199
$endgroup$
$begingroup$
Thank you! For some reason I thought the distribution itself would change, not just the integration limits.
$endgroup$
– 0k33
Dec 9 '18 at 19:09
add a comment |
$begingroup$
Given that you know how to integrate the joint density, then all you have to do is modify the region of integration suitably: The inequality $X < Y + 1$ with the added conditions $X > 0$, $Y > 0$, would represent what region of the $(X,Y)$ coordinate plane? Plot the equation $Y = X - 1$. What does that look like? Now, sketch the region satisfying the inequalities. Then consider how to set up an iterated integral over which you would integrate the joint density: one order of integration will be more inconvenient than the other, so choose carefully.
answered Dec 9 '18 at 17:27
heropupheropup
63k66199
$endgroup$
Given that you know how to integrate the joint density, then all you have to do is modify the region of integration suitably: The inequality $X < Y + 1$ with the added conditions $X > 0$, $Y > 0$, would represent what region of the $(X,Y)$ coordinate plane? Plot the equation $Y = X - 1$. What does that look like? Now, sketch the region satisfying the inequalities. Then consider how to set up an iterated integral over which you would integrate the joint density: one order of integration will be more inconvenient than the other, so choose carefully.
answered Dec 9 '18 at 17:27
heropupheropup
63k66199
answered Dec 9 '18 at 17:27
heropupheropup
63k66199
answered Dec 9 '18 at 17:27
heropupheropup
63k66199
answered Dec 9 '18 at 17:27
heropupheropup
63k66199
63k66199
$begingroup$
Thank you! For some reason I thought the distribution itself would change, not just the integration limits.
$endgroup$
– 0k33
Dec 9 '18 at 19:09
add a comment |
$begingroup$
Thank you! For some reason I thought the distribution itself would change, not just the integration limits.
$endgroup$
– 0k33
Dec 9 '18 at 19:09
$begingroup$
Thank you! For some reason I thought the distribution itself would change, not just the integration limits.
$endgroup$
– 0k33
Dec 9 '18 at 19:09
$begingroup$
Thank you! For some reason I thought the distribution itself would change, not just the integration limits.
$endgroup$
– 0k33
Dec 9 '18 at 19:09
add a comment |
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