Mixing
What is the distribution of $X|W=w$?
$begingroup$
Let $X$ and $Y$ be independent random variables with uniform distribution between $0$ and $1$, that is, have joint density $f_{xy}(x, y) = 1$, if x $in$ $[0,1]$ and y $in [0,1]$ and $f_{xy} (x, y) = 0$, cc. Let $W = (X + Y) / 2$:
What is the distribution of $X|W=w$?
I started considering something like this: $X|W = (x +y)/2$
But I stucked. I dont know how to begin.
Any idea? Hint?
probability probability-theory conditional-probability
asked Jan 14 at 0:31
LinkmanLinkman
1246
$endgroup$
add a comment |
$begingroup$
Let $X$ and $Y$ be independent random variables with uniform distribution between $0$ and $1$, that is, have joint density $f_{xy}(x, y) = 1$, if x $in$ $[0,1]$ and y $in [0,1]$ and $f_{xy} (x, y) = 0$, cc. Let $W = (X + Y) / 2$:
What is the distribution of $X|W=w$?
I started considering something like this: $X|W = (x +y)/2$
But I stucked. I dont know how to begin.
Any idea? Hint?
probability probability-theory conditional-probability
asked Jan 14 at 0:31
LinkmanLinkman
1246
$endgroup$
add a comment |
$begingroup$
Let $X$ and $Y$ be independent random variables with uniform distribution between $0$ and $1$, that is, have joint density $f_{xy}(x, y) = 1$, if x $in$ $[0,1]$ and y $in [0,1]$ and $f_{xy} (x, y) = 0$, cc. Let $W = (X + Y) / 2$:
What is the distribution of $X|W=w$?
I started considering something like this: $X|W = (x +y)/2$
But I stucked. I dont know how to begin.
Any idea? Hint?
probability probability-theory conditional-probability
asked Jan 14 at 0:31
LinkmanLinkman
1246
$endgroup$
Let $X$ and $Y$ be independent random variables with uniform distribution between $0$ and $1$, that is, have joint density $f_{xy}(x, y) = 1$, if x $in$ $[0,1]$ and y $in [0,1]$ and $f_{xy} (x, y) = 0$, cc. Let $W = (X + Y) / 2$:
What is the distribution of $X|W=w$?
I started considering something like this: $X|W = (x +y)/2$
But I stucked. I dont know how to begin.
Any idea? Hint?
probability probability-theory conditional-probability
probability probability-theory conditional-probability
asked Jan 14 at 0:31
LinkmanLinkman
1246
asked Jan 14 at 0:31
LinkmanLinkman
1246
asked Jan 14 at 0:31
LinkmanLinkman
1246
asked Jan 14 at 0:31
LinkmanLinkman
1246
asked Jan 14 at 0:31
LinkmanLinkman
1246
1246
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The conditional pdf is given by
$$f_{X | W = w}(x) =
frac {f_{X, W}(x, w)} {f_W(w)}.$$
$f_W$ is the pdf of a sum of two independent uniformly distributed r.v.:
$$f_W(w) = 4 w left[0 < w leq frac 1 2 right] +
4 (1 - w) left[frac 1 2 < w < 1 right].$$
The transformation $(x,w) = (x, (x + y)/2)$ maps the square $0 < x < 1 land 0 < y < 1$ to the parallelogram $0 < x < 1 land 0 < 2 w - x < 1$ and has the Jacobian $partial(x, w)/partial(x, y) = 1/2$. The transformed pdf is
$$f_{X, W}(x, w) = 2 ,[0 < x < 1 land 0 < 2 w - x < 1] = \
2 left[ 0 < w leq frac 1 2 land 0 < x < 2 w right] +
2 left[ frac 1 2 < w < 1 land 2 w - 1 < x < 1 right].$$
Consider what $f_{X | W = w}$ simplifies to when $w < 1/2$ and when $w > 1/2$.
answered Jan 14 at 8:34
MaximMaxim
5,5981219
$endgroup$
$begingroup$
How did you found out f_{w}? I found 2w.
$endgroup$
– Linkman
Jan 28 at 0:04
1
$begingroup$
Start with finding the cdf: if $0 leq w leq 1/2$, then $operatorname P(Y < 2 w - X)$ requires integrating $f_{X, Y}$ over the triangle with vertices $(0, 0), (2 w, 0), (0, 2w)$, giving $(2 w)^2/2$. (In fact, there is a known formula for an $n$-fold convolution of unit pulses.)
$endgroup$
– Maxim
Jan 28 at 11:03
add a comment |
$begingroup$
$$X = 2W-Y$$
So,
$$X_c = (X|W=w) = (2w-Y_c) $$
Hence, $X_c$ should also be uniform from $[max(2w-1,0),min(2w,1)]$.
EDIT: Here is an argument for why $X_c$ will also be uniform over its domain.
$$P(X=x_1|W=w) = frac{P(X=x_1 & W=w)}{P(W=w)} = frac{P(X=x_1 & Y=2w-x_1)}{P(W=w)}$$
Since $X$ and $Y$ are independent, this becomes:
$$P(X=x_1|W=w) = frac{P(X=x_1)P(Y=2w-x_1)}{P(W=w)}$$
If $2w-x_1 <0 $ or $2w-x_1>1$, the density above becomes $0$.
But otherwise, it will stay exactly the same if we replace $x_1$ by $x_2$ (such that both satisfy $0 leq 2w-x leq 1$). Hence, $X_c$ and $Y_c$ should remain uniform over their valid domains.
answered Jan 14 at 0:38
Rohit PandeyRohit Pandey
1,2871022
$endgroup$
1
$begingroup$
$X_c$ is $X$ conditional on $W=w$. Since by definition, $X+Y=2W$, if you tell me $W=w$, $X_c$ should be $2w-Y$
$endgroup$
– Rohit Pandey
Jan 14 at 0:46
1
$begingroup$
These variables will remain uniform, only their domains will shift (for example, if $W=1$, the domain of $X_c$ becomes $[1,1]$; if $W=0.9$, the domain becomes $[0.8,1]$). Hence, you can calculate anything you like about them, given they are simply uniform over some domain.
$endgroup$
– Rohit Pandey
Jan 14 at 1:09
1
$begingroup$
Really really nice! Thanks!
$endgroup$
– Linkman
Jan 14 at 1:22
1
$begingroup$
@DiogoBastos added explanation for why $X_c$ and $Y_c$ must remain uniform.
$endgroup$
– Rohit Pandey
Jan 14 at 2:09
1
$begingroup$
Just the symmetry between $X$ and $Y$ is not sufficient though. Suppose we had $W = X Y$, then $f_{X | W = w}(x)$ wouldn't be a constant. The difference is that the jacobian of the transformation isn't constant in that case. Also, writing $operatorname P(X = x)$ is problematic, the probability of a continuous r.v. taking a given value is zero.
$endgroup$
– Maxim
Jan 14 at 4:57
|
show 6 more comments
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The conditional pdf is given by
$$f_{X | W = w}(x) =
frac {f_{X, W}(x, w)} {f_W(w)}.$$
$f_W$ is the pdf of a sum of two independent uniformly distributed r.v.:
$$f_W(w) = 4 w left[0 < w leq frac 1 2 right] +
4 (1 - w) left[frac 1 2 < w < 1 right].$$
The transformation $(x,w) = (x, (x + y)/2)$ maps the square $0 < x < 1 land 0 < y < 1$ to the parallelogram $0 < x < 1 land 0 < 2 w - x < 1$ and has the Jacobian $partial(x, w)/partial(x, y) = 1/2$. The transformed pdf is
$$f_{X, W}(x, w) = 2 ,[0 < x < 1 land 0 < 2 w - x < 1] = \
2 left[ 0 < w leq frac 1 2 land 0 < x < 2 w right] +
2 left[ frac 1 2 < w < 1 land 2 w - 1 < x < 1 right].$$
Consider what $f_{X | W = w}$ simplifies to when $w < 1/2$ and when $w > 1/2$.
answered Jan 14 at 8:34
MaximMaxim
5,5981219
$endgroup$
$begingroup$
How did you found out f_{w}? I found 2w.
$endgroup$
– Linkman
Jan 28 at 0:04
1
$begingroup$
Start with finding the cdf: if $0 leq w leq 1/2$, then $operatorname P(Y < 2 w - X)$ requires integrating $f_{X, Y}$ over the triangle with vertices $(0, 0), (2 w, 0), (0, 2w)$, giving $(2 w)^2/2$. (In fact, there is a known formula for an $n$-fold convolution of unit pulses.)
$endgroup$
– Maxim
Jan 28 at 11:03
add a comment |
$begingroup$
The conditional pdf is given by
$$f_{X | W = w}(x) =
frac {f_{X, W}(x, w)} {f_W(w)}.$$
$f_W$ is the pdf of a sum of two independent uniformly distributed r.v.:
$$f_W(w) = 4 w left[0 < w leq frac 1 2 right] +
4 (1 - w) left[frac 1 2 < w < 1 right].$$
The transformation $(x,w) = (x, (x + y)/2)$ maps the square $0 < x < 1 land 0 < y < 1$ to the parallelogram $0 < x < 1 land 0 < 2 w - x < 1$ and has the Jacobian $partial(x, w)/partial(x, y) = 1/2$. The transformed pdf is
$$f_{X, W}(x, w) = 2 ,[0 < x < 1 land 0 < 2 w - x < 1] = \
2 left[ 0 < w leq frac 1 2 land 0 < x < 2 w right] +
2 left[ frac 1 2 < w < 1 land 2 w - 1 < x < 1 right].$$
Consider what $f_{X | W = w}$ simplifies to when $w < 1/2$ and when $w > 1/2$.
answered Jan 14 at 8:34
MaximMaxim
5,5981219
$endgroup$
$begingroup$
How did you found out f_{w}? I found 2w.
$endgroup$
– Linkman
Jan 28 at 0:04
1
$begingroup$
Start with finding the cdf: if $0 leq w leq 1/2$, then $operatorname P(Y < 2 w - X)$ requires integrating $f_{X, Y}$ over the triangle with vertices $(0, 0), (2 w, 0), (0, 2w)$, giving $(2 w)^2/2$. (In fact, there is a known formula for an $n$-fold convolution of unit pulses.)
$endgroup$
– Maxim
Jan 28 at 11:03
add a comment |
$begingroup$
The conditional pdf is given by
$$f_{X | W = w}(x) =
frac {f_{X, W}(x, w)} {f_W(w)}.$$
$f_W$ is the pdf of a sum of two independent uniformly distributed r.v.:
$$f_W(w) = 4 w left[0 < w leq frac 1 2 right] +
4 (1 - w) left[frac 1 2 < w < 1 right].$$
The transformation $(x,w) = (x, (x + y)/2)$ maps the square $0 < x < 1 land 0 < y < 1$ to the parallelogram $0 < x < 1 land 0 < 2 w - x < 1$ and has the Jacobian $partial(x, w)/partial(x, y) = 1/2$. The transformed pdf is
$$f_{X, W}(x, w) = 2 ,[0 < x < 1 land 0 < 2 w - x < 1] = \
2 left[ 0 < w leq frac 1 2 land 0 < x < 2 w right] +
2 left[ frac 1 2 < w < 1 land 2 w - 1 < x < 1 right].$$
Consider what $f_{X | W = w}$ simplifies to when $w < 1/2$ and when $w > 1/2$.
answered Jan 14 at 8:34
MaximMaxim
5,5981219
$endgroup$
The conditional pdf is given by
$$f_{X | W = w}(x) =
frac {f_{X, W}(x, w)} {f_W(w)}.$$
$f_W$ is the pdf of a sum of two independent uniformly distributed r.v.:
$$f_W(w) = 4 w left[0 < w leq frac 1 2 right] +
4 (1 - w) left[frac 1 2 < w < 1 right].$$
The transformation $(x,w) = (x, (x + y)/2)$ maps the square $0 < x < 1 land 0 < y < 1$ to the parallelogram $0 < x < 1 land 0 < 2 w - x < 1$ and has the Jacobian $partial(x, w)/partial(x, y) = 1/2$. The transformed pdf is
$$f_{X, W}(x, w) = 2 ,[0 < x < 1 land 0 < 2 w - x < 1] = \
2 left[ 0 < w leq frac 1 2 land 0 < x < 2 w right] +
2 left[ frac 1 2 < w < 1 land 2 w - 1 < x < 1 right].$$
Consider what $f_{X | W = w}$ simplifies to when $w < 1/2$ and when $w > 1/2$.
answered Jan 14 at 8:34
MaximMaxim
5,5981219
answered Jan 14 at 8:34
MaximMaxim
5,5981219
answered Jan 14 at 8:34
MaximMaxim
5,5981219
answered Jan 14 at 8:34
MaximMaxim
5,5981219
5,5981219
$begingroup$
How did you found out f_{w}? I found 2w.
$endgroup$
– Linkman
Jan 28 at 0:04
1
$begingroup$
Start with finding the cdf: if $0 leq w leq 1/2$, then $operatorname P(Y < 2 w - X)$ requires integrating $f_{X, Y}$ over the triangle with vertices $(0, 0), (2 w, 0), (0, 2w)$, giving $(2 w)^2/2$. (In fact, there is a known formula for an $n$-fold convolution of unit pulses.)
$endgroup$
– Maxim
Jan 28 at 11:03
add a comment |
$begingroup$
How did you found out f_{w}? I found 2w.
$endgroup$
– Linkman
Jan 28 at 0:04
1
$begingroup$
Start with finding the cdf: if $0 leq w leq 1/2$, then $operatorname P(Y < 2 w - X)$ requires integrating $f_{X, Y}$ over the triangle with vertices $(0, 0), (2 w, 0), (0, 2w)$, giving $(2 w)^2/2$. (In fact, there is a known formula for an $n$-fold convolution of unit pulses.)
$endgroup$
– Maxim
Jan 28 at 11:03
$begingroup$
How did you found out f_{w}? I found 2w.
$endgroup$
– Linkman
Jan 28 at 0:04
$begingroup$
How did you found out f_{w}? I found 2w.
$endgroup$
– Linkman
Jan 28 at 0:04
1
1
$begingroup$
Start with finding the cdf: if $0 leq w leq 1/2$, then $operatorname P(Y < 2 w - X)$ requires integrating $f_{X, Y}$ over the triangle with vertices $(0, 0), (2 w, 0), (0, 2w)$, giving $(2 w)^2/2$. (In fact, there is a known formula for an $n$-fold convolution of unit pulses.)
$endgroup$
– Maxim
Jan 28 at 11:03
$begingroup$
Start with finding the cdf: if $0 leq w leq 1/2$, then $operatorname P(Y < 2 w - X)$ requires integrating $f_{X, Y}$ over the triangle with vertices $(0, 0), (2 w, 0), (0, 2w)$, giving $(2 w)^2/2$. (In fact, there is a known formula for an $n$-fold convolution of unit pulses.)
$endgroup$
– Maxim
Jan 28 at 11:03
add a comment |
$begingroup$
$$X = 2W-Y$$
So,
$$X_c = (X|W=w) = (2w-Y_c) $$
Hence, $X_c$ should also be uniform from $[max(2w-1,0),min(2w,1)]$.
EDIT: Here is an argument for why $X_c$ will also be uniform over its domain.
$$P(X=x_1|W=w) = frac{P(X=x_1 & W=w)}{P(W=w)} = frac{P(X=x_1 & Y=2w-x_1)}{P(W=w)}$$
Since $X$ and $Y$ are independent, this becomes:
$$P(X=x_1|W=w) = frac{P(X=x_1)P(Y=2w-x_1)}{P(W=w)}$$
If $2w-x_1 <0 $ or $2w-x_1>1$, the density above becomes $0$.
But otherwise, it will stay exactly the same if we replace $x_1$ by $x_2$ (such that both satisfy $0 leq 2w-x leq 1$). Hence, $X_c$ and $Y_c$ should remain uniform over their valid domains.
answered Jan 14 at 0:38
Rohit PandeyRohit Pandey
1,2871022
$endgroup$
1
$begingroup$
$X_c$ is $X$ conditional on $W=w$. Since by definition, $X+Y=2W$, if you tell me $W=w$, $X_c$ should be $2w-Y$
$endgroup$
– Rohit Pandey
Jan 14 at 0:46
1
$begingroup$
These variables will remain uniform, only their domains will shift (for example, if $W=1$, the domain of $X_c$ becomes $[1,1]$; if $W=0.9$, the domain becomes $[0.8,1]$). Hence, you can calculate anything you like about them, given they are simply uniform over some domain.
$endgroup$
– Rohit Pandey
Jan 14 at 1:09
1
$begingroup$
Really really nice! Thanks!
$endgroup$
– Linkman
Jan 14 at 1:22
1
$begingroup$
@DiogoBastos added explanation for why $X_c$ and $Y_c$ must remain uniform.
$endgroup$
– Rohit Pandey
Jan 14 at 2:09
1
$begingroup$
Just the symmetry between $X$ and $Y$ is not sufficient though. Suppose we had $W = X Y$, then $f_{X | W = w}(x)$ wouldn't be a constant. The difference is that the jacobian of the transformation isn't constant in that case. Also, writing $operatorname P(X = x)$ is problematic, the probability of a continuous r.v. taking a given value is zero.
$endgroup$
– Maxim
Jan 14 at 4:57
|
show 6 more comments
$begingroup$
$$X = 2W-Y$$
So,
$$X_c = (X|W=w) = (2w-Y_c) $$
Hence, $X_c$ should also be uniform from $[max(2w-1,0),min(2w,1)]$.
EDIT: Here is an argument for why $X_c$ will also be uniform over its domain.
$$P(X=x_1|W=w) = frac{P(X=x_1 & W=w)}{P(W=w)} = frac{P(X=x_1 & Y=2w-x_1)}{P(W=w)}$$
Since $X$ and $Y$ are independent, this becomes:
$$P(X=x_1|W=w) = frac{P(X=x_1)P(Y=2w-x_1)}{P(W=w)}$$
If $2w-x_1 <0 $ or $2w-x_1>1$, the density above becomes $0$.
But otherwise, it will stay exactly the same if we replace $x_1$ by $x_2$ (such that both satisfy $0 leq 2w-x leq 1$). Hence, $X_c$ and $Y_c$ should remain uniform over their valid domains.
answered Jan 14 at 0:38
Rohit PandeyRohit Pandey
1,2871022
$endgroup$
1
$begingroup$
$X_c$ is $X$ conditional on $W=w$. Since by definition, $X+Y=2W$, if you tell me $W=w$, $X_c$ should be $2w-Y$
$endgroup$
– Rohit Pandey
Jan 14 at 0:46
1
$begingroup$
These variables will remain uniform, only their domains will shift (for example, if $W=1$, the domain of $X_c$ becomes $[1,1]$; if $W=0.9$, the domain becomes $[0.8,1]$). Hence, you can calculate anything you like about them, given they are simply uniform over some domain.
$endgroup$
– Rohit Pandey
Jan 14 at 1:09
1
$begingroup$
Really really nice! Thanks!
$endgroup$
– Linkman
Jan 14 at 1:22
1
$begingroup$
@DiogoBastos added explanation for why $X_c$ and $Y_c$ must remain uniform.
$endgroup$
– Rohit Pandey
Jan 14 at 2:09
1
$begingroup$
Just the symmetry between $X$ and $Y$ is not sufficient though. Suppose we had $W = X Y$, then $f_{X | W = w}(x)$ wouldn't be a constant. The difference is that the jacobian of the transformation isn't constant in that case. Also, writing $operatorname P(X = x)$ is problematic, the probability of a continuous r.v. taking a given value is zero.
$endgroup$
– Maxim
Jan 14 at 4:57
|
show 6 more comments
$begingroup$
$$X = 2W-Y$$
So,
$$X_c = (X|W=w) = (2w-Y_c) $$
Hence, $X_c$ should also be uniform from $[max(2w-1,0),min(2w,1)]$.
EDIT: Here is an argument for why $X_c$ will also be uniform over its domain.
$$P(X=x_1|W=w) = frac{P(X=x_1 & W=w)}{P(W=w)} = frac{P(X=x_1 & Y=2w-x_1)}{P(W=w)}$$
Since $X$ and $Y$ are independent, this becomes:
$$P(X=x_1|W=w) = frac{P(X=x_1)P(Y=2w-x_1)}{P(W=w)}$$
If $2w-x_1 <0 $ or $2w-x_1>1$, the density above becomes $0$.
But otherwise, it will stay exactly the same if we replace $x_1$ by $x_2$ (such that both satisfy $0 leq 2w-x leq 1$). Hence, $X_c$ and $Y_c$ should remain uniform over their valid domains.
answered Jan 14 at 0:38
Rohit PandeyRohit Pandey
1,2871022
$endgroup$
$$X = 2W-Y$$
So,
$$X_c = (X|W=w) = (2w-Y_c) $$
Hence, $X_c$ should also be uniform from $[max(2w-1,0),min(2w,1)]$.
EDIT: Here is an argument for why $X_c$ will also be uniform over its domain.
$$P(X=x_1|W=w) = frac{P(X=x_1 & W=w)}{P(W=w)} = frac{P(X=x_1 & Y=2w-x_1)}{P(W=w)}$$
Since $X$ and $Y$ are independent, this becomes:
$$P(X=x_1|W=w) = frac{P(X=x_1)P(Y=2w-x_1)}{P(W=w)}$$
If $2w-x_1 <0 $ or $2w-x_1>1$, the density above becomes $0$.
But otherwise, it will stay exactly the same if we replace $x_1$ by $x_2$ (such that both satisfy $0 leq 2w-x leq 1$). Hence, $X_c$ and $Y_c$ should remain uniform over their valid domains.
answered Jan 14 at 0:38
Rohit PandeyRohit Pandey
1,2871022
edited Jan 14 at 5:29
answered Jan 14 at 0:38
Rohit PandeyRohit Pandey
1,2871022
answered Jan 14 at 0:38
Rohit PandeyRohit Pandey
1,2871022
answered Jan 14 at 0:38
Rohit PandeyRohit Pandey
1,2871022
1,2871022
1
$begingroup$
$X_c$ is $X$ conditional on $W=w$. Since by definition, $X+Y=2W$, if you tell me $W=w$, $X_c$ should be $2w-Y$
$endgroup$
– Rohit Pandey
Jan 14 at 0:46
1
$begingroup$
These variables will remain uniform, only their domains will shift (for example, if $W=1$, the domain of $X_c$ becomes $[1,1]$; if $W=0.9$, the domain becomes $[0.8,1]$). Hence, you can calculate anything you like about them, given they are simply uniform over some domain.
$endgroup$
– Rohit Pandey
Jan 14 at 1:09
1
$begingroup$
Really really nice! Thanks!
$endgroup$
– Linkman
Jan 14 at 1:22
1
$begingroup$
@DiogoBastos added explanation for why $X_c$ and $Y_c$ must remain uniform.
$endgroup$
– Rohit Pandey
Jan 14 at 2:09
1
$begingroup$
Just the symmetry between $X$ and $Y$ is not sufficient though. Suppose we had $W = X Y$, then $f_{X | W = w}(x)$ wouldn't be a constant. The difference is that the jacobian of the transformation isn't constant in that case. Also, writing $operatorname P(X = x)$ is problematic, the probability of a continuous r.v. taking a given value is zero.
$endgroup$
– Maxim
Jan 14 at 4:57
|
show 6 more comments
1
$begingroup$
$X_c$ is $X$ conditional on $W=w$. Since by definition, $X+Y=2W$, if you tell me $W=w$, $X_c$ should be $2w-Y$
$endgroup$
– Rohit Pandey
Jan 14 at 0:46
1
$begingroup$
These variables will remain uniform, only their domains will shift (for example, if $W=1$, the domain of $X_c$ becomes $[1,1]$; if $W=0.9$, the domain becomes $[0.8,1]$). Hence, you can calculate anything you like about them, given they are simply uniform over some domain.
$endgroup$
– Rohit Pandey
Jan 14 at 1:09
1
$begingroup$
Really really nice! Thanks!
$endgroup$
– Linkman
Jan 14 at 1:22
1
$begingroup$
@DiogoBastos added explanation for why $X_c$ and $Y_c$ must remain uniform.
$endgroup$
– Rohit Pandey
Jan 14 at 2:09
1
$begingroup$
Just the symmetry between $X$ and $Y$ is not sufficient though. Suppose we had $W = X Y$, then $f_{X | W = w}(x)$ wouldn't be a constant. The difference is that the jacobian of the transformation isn't constant in that case. Also, writing $operatorname P(X = x)$ is problematic, the probability of a continuous r.v. taking a given value is zero.
$endgroup$
– Maxim
Jan 14 at 4:57
1
1
$begingroup$
$X_c$ is $X$ conditional on $W=w$. Since by definition, $X+Y=2W$, if you tell me $W=w$, $X_c$ should be $2w-Y$
$endgroup$
– Rohit Pandey
Jan 14 at 0:46
$begingroup$
$X_c$ is $X$ conditional on $W=w$. Since by definition, $X+Y=2W$, if you tell me $W=w$, $X_c$ should be $2w-Y$
$endgroup$
– Rohit Pandey
Jan 14 at 0:46
1
1
$begingroup$
These variables will remain uniform, only their domains will shift (for example, if $W=1$, the domain of $X_c$ becomes $[1,1]$; if $W=0.9$, the domain becomes $[0.8,1]$). Hence, you can calculate anything you like about them, given they are simply uniform over some domain.
$endgroup$
– Rohit Pandey
Jan 14 at 1:09
$begingroup$
These variables will remain uniform, only their domains will shift (for example, if $W=1$, the domain of $X_c$ becomes $[1,1]$; if $W=0.9$, the domain becomes $[0.8,1]$). Hence, you can calculate anything you like about them, given they are simply uniform over some domain.
$endgroup$
– Rohit Pandey
Jan 14 at 1:09
1
1
$begingroup$
Really really nice! Thanks!
$endgroup$
– Linkman
Jan 14 at 1:22
$begingroup$
Really really nice! Thanks!
$endgroup$
– Linkman
Jan 14 at 1:22
1
1
$begingroup$
@DiogoBastos added explanation for why $X_c$ and $Y_c$ must remain uniform.
$endgroup$
– Rohit Pandey
Jan 14 at 2:09
$begingroup$
@DiogoBastos added explanation for why $X_c$ and $Y_c$ must remain uniform.
$endgroup$
– Rohit Pandey
Jan 14 at 2:09
1
1
$begingroup$
Just the symmetry between $X$ and $Y$ is not sufficient though. Suppose we had $W = X Y$, then $f_{X | W = w}(x)$ wouldn't be a constant. The difference is that the jacobian of the transformation isn't constant in that case. Also, writing $operatorname P(X = x)$ is problematic, the probability of a continuous r.v. taking a given value is zero.
$endgroup$
– Maxim
Jan 14 at 4:57
$begingroup$
Just the symmetry between $X$ and $Y$ is not sufficient though. Suppose we had $W = X Y$, then $f_{X | W = w}(x)$ wouldn't be a constant. The difference is that the jacobian of the transformation isn't constant in that case. Also, writing $operatorname P(X = x)$ is problematic, the probability of a continuous r.v. taking a given value is zero.
$endgroup$
– Maxim
Jan 14 at 4:57
|
show 6 more comments
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