Mixing
Probability: Conditional expectation from a joint pdf
$begingroup$
Let $X$ and $Y$ have the joint pdf $f(x;y)=2exp(x+y)$, $0 < x < y < infty$, zero elsewhere.
Find the conditional mean $E(Y|X=x)$.
This seems like a simple problem. I know I have to find the marginal pdf of $x$ and then divide the joint pdf by the marginal pdf. But the marginal pdf of $X$ diverges, so I don't know how to proceed.
probability
edited Jan 21 at 11:14
Henrik
6,03592030
asked May 6 '18 at 0:50
KellyKelly
236
$endgroup$
add a comment |
$begingroup$
Let $X$ and $Y$ have the joint pdf $f(x;y)=2exp(x+y)$, $0 < x < y < infty$, zero elsewhere.
Find the conditional mean $E(Y|X=x)$.
This seems like a simple problem. I know I have to find the marginal pdf of $x$ and then divide the joint pdf by the marginal pdf. But the marginal pdf of $X$ diverges, so I don't know how to proceed.
probability
edited Jan 21 at 11:14
Henrik
6,03592030
asked May 6 '18 at 0:50
KellyKelly
236
$endgroup$
add a comment |
$begingroup$
Let $X$ and $Y$ have the joint pdf $f(x;y)=2exp(x+y)$, $0 < x < y < infty$, zero elsewhere.
Find the conditional mean $E(Y|X=x)$.
This seems like a simple problem. I know I have to find the marginal pdf of $x$ and then divide the joint pdf by the marginal pdf. But the marginal pdf of $X$ diverges, so I don't know how to proceed.
probability
edited Jan 21 at 11:14
Henrik
6,03592030
asked May 6 '18 at 0:50
KellyKelly
236
$endgroup$
Let $X$ and $Y$ have the joint pdf $f(x;y)=2exp(x+y)$, $0 < x < y < infty$, zero elsewhere.
Find the conditional mean $E(Y|X=x)$.
This seems like a simple problem. I know I have to find the marginal pdf of $x$ and then divide the joint pdf by the marginal pdf. But the marginal pdf of $X$ diverges, so I don't know how to proceed.
probability
probability
edited Jan 21 at 11:14
Henrik
6,03592030
asked May 6 '18 at 0:50
KellyKelly
236
edited Jan 21 at 11:14
Henrik
6,03592030
asked May 6 '18 at 0:50
KellyKelly
236
edited Jan 21 at 11:14
Henrik
6,03592030
edited Jan 21 at 11:14
Henrik
6,03592030
edited Jan 21 at 11:14
Henrik
6,03592030
6,03592030
asked May 6 '18 at 0:50
KellyKelly
236
asked May 6 '18 at 0:50
KellyKelly
236
asked May 6 '18 at 0:50
KellyKelly
236
236
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
There is probably a mistake in the question.
In fact, note that
$$int_0^infty int_x^infty 2exp(x+y) ,, dy dx ne 1.$$
If we fix an $x$, $y$ can get arbitrarily large, and the so call pdf can get arbitraily large and it can't converges.
We do have
$$int_0^infty int_x^infty 2exp(color{red}-(x+y)) ,, dy dx = 1.$$
answered May 6 '18 at 1:05
Siong Thye GohSiong Thye Goh
102k1467119
$endgroup$
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
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active
oldest
votes
$begingroup$
There is probably a mistake in the question.
In fact, note that
$$int_0^infty int_x^infty 2exp(x+y) ,, dy dx ne 1.$$
If we fix an $x$, $y$ can get arbitrarily large, and the so call pdf can get arbitraily large and it can't converges.
We do have
$$int_0^infty int_x^infty 2exp(color{red}-(x+y)) ,, dy dx = 1.$$
answered May 6 '18 at 1:05
Siong Thye GohSiong Thye Goh
102k1467119
$endgroup$
add a comment |
$begingroup$
There is probably a mistake in the question.
In fact, note that
$$int_0^infty int_x^infty 2exp(x+y) ,, dy dx ne 1.$$
If we fix an $x$, $y$ can get arbitrarily large, and the so call pdf can get arbitraily large and it can't converges.
We do have
$$int_0^infty int_x^infty 2exp(color{red}-(x+y)) ,, dy dx = 1.$$
answered May 6 '18 at 1:05
Siong Thye GohSiong Thye Goh
102k1467119
$endgroup$
add a comment |
$begingroup$
There is probably a mistake in the question.
In fact, note that
$$int_0^infty int_x^infty 2exp(x+y) ,, dy dx ne 1.$$
If we fix an $x$, $y$ can get arbitrarily large, and the so call pdf can get arbitraily large and it can't converges.
We do have
$$int_0^infty int_x^infty 2exp(color{red}-(x+y)) ,, dy dx = 1.$$
answered May 6 '18 at 1:05
Siong Thye GohSiong Thye Goh
102k1467119
$endgroup$
There is probably a mistake in the question.
In fact, note that
$$int_0^infty int_x^infty 2exp(x+y) ,, dy dx ne 1.$$
If we fix an $x$, $y$ can get arbitrarily large, and the so call pdf can get arbitraily large and it can't converges.
We do have
$$int_0^infty int_x^infty 2exp(color{red}-(x+y)) ,, dy dx = 1.$$
answered May 6 '18 at 1:05
Siong Thye GohSiong Thye Goh
102k1467119
answered May 6 '18 at 1:05
Siong Thye GohSiong Thye Goh
102k1467119
answered May 6 '18 at 1:05
Siong Thye GohSiong Thye Goh
102k1467119
answered May 6 '18 at 1:05
Siong Thye GohSiong Thye Goh
102k1467119
102k1467119
add a comment |
add a comment |
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