Given Two-dimensional Gaussian distribution, how to derive the pdf of Z?
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0
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Let $(X,Y) sim N left(
begin{pmatrix}0\0
end{pmatrix},
begin{pmatrix} 1& 0 \
0 & 1
end{pmatrix}
right)$ and $Z = X I {XY >0} - X I {XY <0} $
Find the distribution of Z.
From the question, I only know that X,Y are independent and X~N(0,1), Y~N(0,1). I think we should derive the cdf of Z for z>0 and z<0 first in order to get the pdf of Z. However, I don't know how to, since Z=X for XY>0 and Z=-X for XY<0, which is really complicate. Somebody help me please!
probability statistics
edited Nov 16 at 21:49
leonbloy
39.7k645106
asked Nov 16 at 21:42
Araceli
61
add a comment |
up vote
0
down vote
favorite
Let $(X,Y) sim N left(
begin{pmatrix}0\0
end{pmatrix},
begin{pmatrix} 1& 0 \
0 & 1
end{pmatrix}
right)$ and $Z = X I {XY >0} - X I {XY <0} $
Find the distribution of Z.
From the question, I only know that X,Y are independent and X~N(0,1), Y~N(0,1). I think we should derive the cdf of Z for z>0 and z<0 first in order to get the pdf of Z. However, I don't know how to, since Z=X for XY>0 and Z=-X for XY<0, which is really complicate. Somebody help me please!
probability statistics
edited Nov 16 at 21:49
leonbloy
39.7k645106
asked Nov 16 at 21:42
Araceli
61
Hi Araceli, welcome to MSE. I wrote down the problem statement in Latex (as you should here), please check that it's ok.
– leonbloy
Nov 16 at 21:50
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $(X,Y) sim N left(
begin{pmatrix}0\0
end{pmatrix},
begin{pmatrix} 1& 0 \
0 & 1
end{pmatrix}
right)$ and $Z = X I {XY >0} - X I {XY <0} $
Find the distribution of Z.
From the question, I only know that X,Y are independent and X~N(0,1), Y~N(0,1). I think we should derive the cdf of Z for z>0 and z<0 first in order to get the pdf of Z. However, I don't know how to, since Z=X for XY>0 and Z=-X for XY<0, which is really complicate. Somebody help me please!
probability statistics
edited Nov 16 at 21:49
leonbloy
39.7k645106
asked Nov 16 at 21:42
Araceli
61
Let $(X,Y) sim N left(
begin{pmatrix}0\0
end{pmatrix},
begin{pmatrix} 1& 0 \
0 & 1
end{pmatrix}
right)$ and $Z = X I {XY >0} - X I {XY <0} $
Find the distribution of Z.
From the question, I only know that X,Y are independent and X~N(0,1), Y~N(0,1). I think we should derive the cdf of Z for z>0 and z<0 first in order to get the pdf of Z. However, I don't know how to, since Z=X for XY>0 and Z=-X for XY<0, which is really complicate. Somebody help me please!
probability statistics
probability statistics
edited Nov 16 at 21:49
leonbloy
39.7k645106
asked Nov 16 at 21:42
Araceli
61
edited Nov 16 at 21:49
leonbloy
39.7k645106
asked Nov 16 at 21:42
Araceli
61
edited Nov 16 at 21:49
leonbloy
39.7k645106
edited Nov 16 at 21:49
leonbloy
39.7k645106
edited Nov 16 at 21:49
leonbloy
39.7k645106
39.7k645106
asked Nov 16 at 21:42
Araceli
61
asked Nov 16 at 21:42
Araceli
61
asked Nov 16 at 21:42
Araceli
61
61
Hi Araceli, welcome to MSE. I wrote down the problem statement in Latex (as you should here), please check that it's ok.
– leonbloy
Nov 16 at 21:50
add a comment |
Hi Araceli, welcome to MSE. I wrote down the problem statement in Latex (as you should here), please check that it's ok.
– leonbloy
Nov 16 at 21:50
Hi Araceli, welcome to MSE. I wrote down the problem statement in Latex (as you should here), please check that it's ok.
– leonbloy
Nov 16 at 21:50
Hi Araceli, welcome to MSE. I wrote down the problem statement in Latex (as you should here), please check that it's ok.
– leonbloy
Nov 16 at 21:50
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
There are several ways...
For example : Let $W=I{X Y ge 0 }$. Notice (show) that $P(W=0) = P(W=1)=frac12$ and (important, and not so trivial) $W$ is independent of $X$.
$$begin{align}P(Zle z) &= P(Z le z cap W=0) + P(Z le z cap W=1) \
&= P(Z le z mid W=0) P(W=0) + P(Z le z mid W=1)P(W=1)\
&= P(X le z mid W=0) frac12 + P(-X le z mid W=1)frac12\
&= P(X le z ) frac12 + P(X ge -z )frac12\
&= frac12 ( F_X(z ) + (1- F_X(-z))\
&= F_X(z )
end{align}$$
That $W$ is independent of $X$ is intuitive by symmetry. A little more formally:
$$P(W=1 | X=x) =begin{cases}
P(Yge 0)=frac12 & {rm if hskip 3mm} x ge 0\
P(Y<0)=frac12 & {rm if hskip 3mm} x < 0\
end{cases}$$
Hence $P(W | X) = P(W)$ , and $W,X$ are independent.
answered Nov 16 at 22:03
leonbloy
39.7k645106
Thanks for answering! That's very helpful. However, I'm still a little bit confused about whether W and X are independent. My thought is that whether xy are positive or negative has nothing to do with the distribution of f(x). Do you have a better explanation?
– Araceli
Nov 16 at 23:54
@Araceli I added a possible explanation
– leonbloy
Nov 17 at 2:44
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
There are several ways...
For example : Let $W=I{X Y ge 0 }$. Notice (show) that $P(W=0) = P(W=1)=frac12$ and (important, and not so trivial) $W$ is independent of $X$.
$$begin{align}P(Zle z) &= P(Z le z cap W=0) + P(Z le z cap W=1) \
&= P(Z le z mid W=0) P(W=0) + P(Z le z mid W=1)P(W=1)\
&= P(X le z mid W=0) frac12 + P(-X le z mid W=1)frac12\
&= P(X le z ) frac12 + P(X ge -z )frac12\
&= frac12 ( F_X(z ) + (1- F_X(-z))\
&= F_X(z )
end{align}$$
That $W$ is independent of $X$ is intuitive by symmetry. A little more formally:
$$P(W=1 | X=x) =begin{cases}
P(Yge 0)=frac12 & {rm if hskip 3mm} x ge 0\
P(Y<0)=frac12 & {rm if hskip 3mm} x < 0\
end{cases}$$
Hence $P(W | X) = P(W)$ , and $W,X$ are independent.
answered Nov 16 at 22:03
leonbloy
39.7k645106
Thanks for answering! That's very helpful. However, I'm still a little bit confused about whether W and X are independent. My thought is that whether xy are positive or negative has nothing to do with the distribution of f(x). Do you have a better explanation?
– Araceli
Nov 16 at 23:54
@Araceli I added a possible explanation
– leonbloy
Nov 17 at 2:44
add a comment |
up vote
0
down vote
accepted
There are several ways...
For example : Let $W=I{X Y ge 0 }$. Notice (show) that $P(W=0) = P(W=1)=frac12$ and (important, and not so trivial) $W$ is independent of $X$.
$$begin{align}P(Zle z) &= P(Z le z cap W=0) + P(Z le z cap W=1) \
&= P(Z le z mid W=0) P(W=0) + P(Z le z mid W=1)P(W=1)\
&= P(X le z mid W=0) frac12 + P(-X le z mid W=1)frac12\
&= P(X le z ) frac12 + P(X ge -z )frac12\
&= frac12 ( F_X(z ) + (1- F_X(-z))\
&= F_X(z )
end{align}$$
That $W$ is independent of $X$ is intuitive by symmetry. A little more formally:
$$P(W=1 | X=x) =begin{cases}
P(Yge 0)=frac12 & {rm if hskip 3mm} x ge 0\
P(Y<0)=frac12 & {rm if hskip 3mm} x < 0\
end{cases}$$
Hence $P(W | X) = P(W)$ , and $W,X$ are independent.
answered Nov 16 at 22:03
leonbloy
39.7k645106
Thanks for answering! That's very helpful. However, I'm still a little bit confused about whether W and X are independent. My thought is that whether xy are positive or negative has nothing to do with the distribution of f(x). Do you have a better explanation?
– Araceli
Nov 16 at 23:54
@Araceli I added a possible explanation
– leonbloy
Nov 17 at 2:44
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
There are several ways...
For example : Let $W=I{X Y ge 0 }$. Notice (show) that $P(W=0) = P(W=1)=frac12$ and (important, and not so trivial) $W$ is independent of $X$.
$$begin{align}P(Zle z) &= P(Z le z cap W=0) + P(Z le z cap W=1) \
&= P(Z le z mid W=0) P(W=0) + P(Z le z mid W=1)P(W=1)\
&= P(X le z mid W=0) frac12 + P(-X le z mid W=1)frac12\
&= P(X le z ) frac12 + P(X ge -z )frac12\
&= frac12 ( F_X(z ) + (1- F_X(-z))\
&= F_X(z )
end{align}$$
That $W$ is independent of $X$ is intuitive by symmetry. A little more formally:
$$P(W=1 | X=x) =begin{cases}
P(Yge 0)=frac12 & {rm if hskip 3mm} x ge 0\
P(Y<0)=frac12 & {rm if hskip 3mm} x < 0\
end{cases}$$
Hence $P(W | X) = P(W)$ , and $W,X$ are independent.
answered Nov 16 at 22:03
leonbloy
39.7k645106
There are several ways...
For example : Let $W=I{X Y ge 0 }$. Notice (show) that $P(W=0) = P(W=1)=frac12$ and (important, and not so trivial) $W$ is independent of $X$.
$$begin{align}P(Zle z) &= P(Z le z cap W=0) + P(Z le z cap W=1) \
&= P(Z le z mid W=0) P(W=0) + P(Z le z mid W=1)P(W=1)\
&= P(X le z mid W=0) frac12 + P(-X le z mid W=1)frac12\
&= P(X le z ) frac12 + P(X ge -z )frac12\
&= frac12 ( F_X(z ) + (1- F_X(-z))\
&= F_X(z )
end{align}$$
That $W$ is independent of $X$ is intuitive by symmetry. A little more formally:
$$P(W=1 | X=x) =begin{cases}
P(Yge 0)=frac12 & {rm if hskip 3mm} x ge 0\
P(Y<0)=frac12 & {rm if hskip 3mm} x < 0\
end{cases}$$
Hence $P(W | X) = P(W)$ , and $W,X$ are independent.
answered Nov 16 at 22:03
leonbloy
39.7k645106
edited Nov 17 at 2:43
answered Nov 16 at 22:03
leonbloy
39.7k645106
answered Nov 16 at 22:03
leonbloy
39.7k645106
answered Nov 16 at 22:03
leonbloy
39.7k645106
39.7k645106
Thanks for answering! That's very helpful. However, I'm still a little bit confused about whether W and X are independent. My thought is that whether xy are positive or negative has nothing to do with the distribution of f(x). Do you have a better explanation?
– Araceli
Nov 16 at 23:54
@Araceli I added a possible explanation
– leonbloy
Nov 17 at 2:44
add a comment |
Thanks for answering! That's very helpful. However, I'm still a little bit confused about whether W and X are independent. My thought is that whether xy are positive or negative has nothing to do with the distribution of f(x). Do you have a better explanation?
– Araceli
Nov 16 at 23:54
@Araceli I added a possible explanation
– leonbloy
Nov 17 at 2:44
Thanks for answering! That's very helpful. However, I'm still a little bit confused about whether W and X are independent. My thought is that whether xy are positive or negative has nothing to do with the distribution of f(x). Do you have a better explanation?
– Araceli
Nov 16 at 23:54
Thanks for answering! That's very helpful. However, I'm still a little bit confused about whether W and X are independent. My thought is that whether xy are positive or negative has nothing to do with the distribution of f(x). Do you have a better explanation?
– Araceli
Nov 16 at 23:54
@Araceli I added a possible explanation
– leonbloy
Nov 17 at 2:44
@Araceli I added a possible explanation
– leonbloy
Nov 17 at 2:44
add a comment |
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Hi Araceli, welcome to MSE. I wrote down the problem statement in Latex (as you should here), please check that it's ok.
– leonbloy
Nov 16 at 21:50