Mixing
Time transformation of random variable with mass points
$begingroup$
Let $B=left{bright}$ denote the set of atoms of the distribution function G. Define the quantile function $G^{-1}left( aright) = inf left{ x in R : G(x) ge a right}$.
Let V be independent of $X$ and uniformly distributed on [0,1]. Set $$
U=left{
begin{array}{cc}
Gleft( Xright) & if~Xnotin B \
Gleft( b-right) +Gleft{ bright} V & if~X=bin B%
end{array}%
right. ,
$$
where $Gleft{ bright} = Gleft( bright) - Gleft( b-right)$.
Then, $U$ is uniformly distributed on $left[ 0,1right] $ and $%
X=G^{-1}left( Uright)$.
I am having a hard time proving this result. Can anyone provide any help?
probability statistics probability-distributions random-variables transformation
asked Jan 18 at 6:15
user17645user17645
32
$endgroup$
add a comment |
$begingroup$
Let $B=left{bright}$ denote the set of atoms of the distribution function G. Define the quantile function $G^{-1}left( aright) = inf left{ x in R : G(x) ge a right}$.
Let V be independent of $X$ and uniformly distributed on [0,1]. Set $$
U=left{
begin{array}{cc}
Gleft( Xright) & if~Xnotin B \
Gleft( b-right) +Gleft{ bright} V & if~X=bin B%
end{array}%
right. ,
$$
where $Gleft{ bright} = Gleft( bright) - Gleft( b-right)$.
Then, $U$ is uniformly distributed on $left[ 0,1right] $ and $%
X=G^{-1}left( Uright)$.
I am having a hard time proving this result. Can anyone provide any help?
probability statistics probability-distributions random-variables transformation
asked Jan 18 at 6:15
user17645user17645
32
$endgroup$
add a comment |
$begingroup$
Let $B=left{bright}$ denote the set of atoms of the distribution function G. Define the quantile function $G^{-1}left( aright) = inf left{ x in R : G(x) ge a right}$.
Let V be independent of $X$ and uniformly distributed on [0,1]. Set $$
U=left{
begin{array}{cc}
Gleft( Xright) & if~Xnotin B \
Gleft( b-right) +Gleft{ bright} V & if~X=bin B%
end{array}%
right. ,
$$
where $Gleft{ bright} = Gleft( bright) - Gleft( b-right)$.
Then, $U$ is uniformly distributed on $left[ 0,1right] $ and $%
X=G^{-1}left( Uright)$.
I am having a hard time proving this result. Can anyone provide any help?
probability statistics probability-distributions random-variables transformation
asked Jan 18 at 6:15
user17645user17645
32
$endgroup$
Let $B=left{bright}$ denote the set of atoms of the distribution function G. Define the quantile function $G^{-1}left( aright) = inf left{ x in R : G(x) ge a right}$.
Let V be independent of $X$ and uniformly distributed on [0,1]. Set $$
U=left{
begin{array}{cc}
Gleft( Xright) & if~Xnotin B \
Gleft( b-right) +Gleft{ bright} V & if~X=bin B%
end{array}%
right. ,
$$
where $Gleft{ bright} = Gleft( bright) - Gleft( b-right)$.
Then, $U$ is uniformly distributed on $left[ 0,1right] $ and $%
X=G^{-1}left( Uright)$.
I am having a hard time proving this result. Can anyone provide any help?
probability statistics probability-distributions random-variables transformation
probability statistics probability-distributions random-variables transformation
asked Jan 18 at 6:15
user17645user17645
32
asked Jan 18 at 6:15
user17645user17645
32
edited Jan 18 at 16:56
user17645
asked Jan 18 at 6:15
user17645user17645
32
asked Jan 18 at 6:15
user17645user17645
32
asked Jan 18 at 6:15
user17645user17645
32
32
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Note that if $X=bin B$ and $b<G^{-1}(u)$, then
$$
U=G(b^-)+G{b}Vle G(b)<u.
$$ Hence $Uge u$ implies that $bge G^{-1}(u)=:w$. For each $uin (0,1)$, we will calculate $P(Uge u)$ dividing it into two cases.
Firstly, if it holds $G{w}=0$, then $Xin B$ and $X=bge w$ implies $Uge G(b^-)ge G(w^-)=G(w)=u.$ Also by the first proposition, $Uge u$ implies $bge w$. If $Xnotin B$ then $U=G(X)ge u$ if and only if $Xge w$. Hence we have
$$begin{eqnarray}
P(Uge u)&=&P(Uge u, Xnotin B)+P(Uge u,Xin B)\
&=&P(Xge w, Xnotin B) +P(Xge w, Xin B)=P(Xge w)=1-G(w^-)=1-u.
end{eqnarray}$$
Secondly, assume $G{w}>0$ and $Uge u$. In case $Xnotin B$, $U=G(X)ge u$ is equivalent to $Xge w$. In case $Xin B$, we see that $X=b>w$ implies $Uge G(b^-)ge G(w)ge u$. A more subtle case is when $X=win B$. Then $U=G(w^-)+G{w}Vge u$ if and only if $Vge frac{u-G(w^-)}{G{w}}$. Summing this up, we have
$$begin{eqnarray}
P(Uge u)&=&P(Uge u, Xnotin B)+P(Uge u,Xin B)\
&=&P(Xge w, Xnotin B) +P(X> w, Xin B)+P(X= w, Vge frac{u-G(w^-)}{G{w}})\
&=&P(Xge w, Xnotin B) +P(X> w, Xin B)+P(X= w) left(1-frac{u-G(w^-)}{G{w}}right)\
&=&P(X>w) + G{w}left(1-frac{u-G(w^-)}{G{w}}right)\
&=&1-G(w)+G{w}-u+G(w^-)=1-u.
end{eqnarray}$$
Thus, it follows that $U$ is uniformly distributed on $[0,1]$.
answered Jan 18 at 20:03
SongSong
15.4k1736
$endgroup$
$begingroup$
Thanks a lot! Any hint on the quantile?
$endgroup$
– user17645
Jan 18 at 20:25
$begingroup$
The quantile function is as you defined. It has properties such as $$G^{-1}(a)=color{red} min{xinBbb R:G(x)ge a}$$ and $$G(x)ge a Longleftrightarrow xge G^{-1}(a).$$
$endgroup$
– Song
Jan 18 at 20:30
$begingroup$
I am having a very hard time using the properties you mentioned to prove that $X =G^{-1}(U)$ a.s.. Would you mind providing additional detailed explanation? Thanks!!!
$endgroup$
– user17645
Jan 22 at 8:23
$begingroup$
@user17645 I just realized that this was a part of your question ... Sorry for that. Now, to show $X=G^{-1}(U)$, first notice that $Ule G(X)$ with probability $1$. In view of the definition of $G^{-1}$, now it suffices to show that $$ x<X implies G(x)<U.$$ Analyze each case of $Xin B$ and $Xnotin B$. It would not be so difficult.
$endgroup$
– Song
Jan 22 at 16:38
add a comment |
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$begingroup$
Note that if $X=bin B$ and $b<G^{-1}(u)$, then
$$
U=G(b^-)+G{b}Vle G(b)<u.
$$ Hence $Uge u$ implies that $bge G^{-1}(u)=:w$. For each $uin (0,1)$, we will calculate $P(Uge u)$ dividing it into two cases.
Firstly, if it holds $G{w}=0$, then $Xin B$ and $X=bge w$ implies $Uge G(b^-)ge G(w^-)=G(w)=u.$ Also by the first proposition, $Uge u$ implies $bge w$. If $Xnotin B$ then $U=G(X)ge u$ if and only if $Xge w$. Hence we have
$$begin{eqnarray}
P(Uge u)&=&P(Uge u, Xnotin B)+P(Uge u,Xin B)\
&=&P(Xge w, Xnotin B) +P(Xge w, Xin B)=P(Xge w)=1-G(w^-)=1-u.
end{eqnarray}$$
Secondly, assume $G{w}>0$ and $Uge u$. In case $Xnotin B$, $U=G(X)ge u$ is equivalent to $Xge w$. In case $Xin B$, we see that $X=b>w$ implies $Uge G(b^-)ge G(w)ge u$. A more subtle case is when $X=win B$. Then $U=G(w^-)+G{w}Vge u$ if and only if $Vge frac{u-G(w^-)}{G{w}}$. Summing this up, we have
$$begin{eqnarray}
P(Uge u)&=&P(Uge u, Xnotin B)+P(Uge u,Xin B)\
&=&P(Xge w, Xnotin B) +P(X> w, Xin B)+P(X= w, Vge frac{u-G(w^-)}{G{w}})\
&=&P(Xge w, Xnotin B) +P(X> w, Xin B)+P(X= w) left(1-frac{u-G(w^-)}{G{w}}right)\
&=&P(X>w) + G{w}left(1-frac{u-G(w^-)}{G{w}}right)\
&=&1-G(w)+G{w}-u+G(w^-)=1-u.
end{eqnarray}$$
Thus, it follows that $U$ is uniformly distributed on $[0,1]$.
answered Jan 18 at 20:03
SongSong
15.4k1736
$endgroup$
$begingroup$
Thanks a lot! Any hint on the quantile?
$endgroup$
– user17645
Jan 18 at 20:25
$begingroup$
The quantile function is as you defined. It has properties such as $$G^{-1}(a)=color{red} min{xinBbb R:G(x)ge a}$$ and $$G(x)ge a Longleftrightarrow xge G^{-1}(a).$$
$endgroup$
– Song
Jan 18 at 20:30
$begingroup$
I am having a very hard time using the properties you mentioned to prove that $X =G^{-1}(U)$ a.s.. Would you mind providing additional detailed explanation? Thanks!!!
$endgroup$
– user17645
Jan 22 at 8:23
$begingroup$
@user17645 I just realized that this was a part of your question ... Sorry for that. Now, to show $X=G^{-1}(U)$, first notice that $Ule G(X)$ with probability $1$. In view of the definition of $G^{-1}$, now it suffices to show that $$ x<X implies G(x)<U.$$ Analyze each case of $Xin B$ and $Xnotin B$. It would not be so difficult.
$endgroup$
– Song
Jan 22 at 16:38
add a comment |
$begingroup$
Note that if $X=bin B$ and $b<G^{-1}(u)$, then
$$
U=G(b^-)+G{b}Vle G(b)<u.
$$ Hence $Uge u$ implies that $bge G^{-1}(u)=:w$. For each $uin (0,1)$, we will calculate $P(Uge u)$ dividing it into two cases.
Firstly, if it holds $G{w}=0$, then $Xin B$ and $X=bge w$ implies $Uge G(b^-)ge G(w^-)=G(w)=u.$ Also by the first proposition, $Uge u$ implies $bge w$. If $Xnotin B$ then $U=G(X)ge u$ if and only if $Xge w$. Hence we have
$$begin{eqnarray}
P(Uge u)&=&P(Uge u, Xnotin B)+P(Uge u,Xin B)\
&=&P(Xge w, Xnotin B) +P(Xge w, Xin B)=P(Xge w)=1-G(w^-)=1-u.
end{eqnarray}$$
Secondly, assume $G{w}>0$ and $Uge u$. In case $Xnotin B$, $U=G(X)ge u$ is equivalent to $Xge w$. In case $Xin B$, we see that $X=b>w$ implies $Uge G(b^-)ge G(w)ge u$. A more subtle case is when $X=win B$. Then $U=G(w^-)+G{w}Vge u$ if and only if $Vge frac{u-G(w^-)}{G{w}}$. Summing this up, we have
$$begin{eqnarray}
P(Uge u)&=&P(Uge u, Xnotin B)+P(Uge u,Xin B)\
&=&P(Xge w, Xnotin B) +P(X> w, Xin B)+P(X= w, Vge frac{u-G(w^-)}{G{w}})\
&=&P(Xge w, Xnotin B) +P(X> w, Xin B)+P(X= w) left(1-frac{u-G(w^-)}{G{w}}right)\
&=&P(X>w) + G{w}left(1-frac{u-G(w^-)}{G{w}}right)\
&=&1-G(w)+G{w}-u+G(w^-)=1-u.
end{eqnarray}$$
Thus, it follows that $U$ is uniformly distributed on $[0,1]$.
answered Jan 18 at 20:03
SongSong
15.4k1736
$endgroup$
$begingroup$
Thanks a lot! Any hint on the quantile?
$endgroup$
– user17645
Jan 18 at 20:25
$begingroup$
The quantile function is as you defined. It has properties such as $$G^{-1}(a)=color{red} min{xinBbb R:G(x)ge a}$$ and $$G(x)ge a Longleftrightarrow xge G^{-1}(a).$$
$endgroup$
– Song
Jan 18 at 20:30
$begingroup$
I am having a very hard time using the properties you mentioned to prove that $X =G^{-1}(U)$ a.s.. Would you mind providing additional detailed explanation? Thanks!!!
$endgroup$
– user17645
Jan 22 at 8:23
$begingroup$
@user17645 I just realized that this was a part of your question ... Sorry for that. Now, to show $X=G^{-1}(U)$, first notice that $Ule G(X)$ with probability $1$. In view of the definition of $G^{-1}$, now it suffices to show that $$ x<X implies G(x)<U.$$ Analyze each case of $Xin B$ and $Xnotin B$. It would not be so difficult.
$endgroup$
– Song
Jan 22 at 16:38
add a comment |
$begingroup$
Note that if $X=bin B$ and $b<G^{-1}(u)$, then
$$
U=G(b^-)+G{b}Vle G(b)<u.
$$ Hence $Uge u$ implies that $bge G^{-1}(u)=:w$. For each $uin (0,1)$, we will calculate $P(Uge u)$ dividing it into two cases.
Firstly, if it holds $G{w}=0$, then $Xin B$ and $X=bge w$ implies $Uge G(b^-)ge G(w^-)=G(w)=u.$ Also by the first proposition, $Uge u$ implies $bge w$. If $Xnotin B$ then $U=G(X)ge u$ if and only if $Xge w$. Hence we have
$$begin{eqnarray}
P(Uge u)&=&P(Uge u, Xnotin B)+P(Uge u,Xin B)\
&=&P(Xge w, Xnotin B) +P(Xge w, Xin B)=P(Xge w)=1-G(w^-)=1-u.
end{eqnarray}$$
Secondly, assume $G{w}>0$ and $Uge u$. In case $Xnotin B$, $U=G(X)ge u$ is equivalent to $Xge w$. In case $Xin B$, we see that $X=b>w$ implies $Uge G(b^-)ge G(w)ge u$. A more subtle case is when $X=win B$. Then $U=G(w^-)+G{w}Vge u$ if and only if $Vge frac{u-G(w^-)}{G{w}}$. Summing this up, we have
$$begin{eqnarray}
P(Uge u)&=&P(Uge u, Xnotin B)+P(Uge u,Xin B)\
&=&P(Xge w, Xnotin B) +P(X> w, Xin B)+P(X= w, Vge frac{u-G(w^-)}{G{w}})\
&=&P(Xge w, Xnotin B) +P(X> w, Xin B)+P(X= w) left(1-frac{u-G(w^-)}{G{w}}right)\
&=&P(X>w) + G{w}left(1-frac{u-G(w^-)}{G{w}}right)\
&=&1-G(w)+G{w}-u+G(w^-)=1-u.
end{eqnarray}$$
Thus, it follows that $U$ is uniformly distributed on $[0,1]$.
answered Jan 18 at 20:03
SongSong
15.4k1736
$endgroup$
Note that if $X=bin B$ and $b<G^{-1}(u)$, then
$$
U=G(b^-)+G{b}Vle G(b)<u.
$$ Hence $Uge u$ implies that $bge G^{-1}(u)=:w$. For each $uin (0,1)$, we will calculate $P(Uge u)$ dividing it into two cases.
Firstly, if it holds $G{w}=0$, then $Xin B$ and $X=bge w$ implies $Uge G(b^-)ge G(w^-)=G(w)=u.$ Also by the first proposition, $Uge u$ implies $bge w$. If $Xnotin B$ then $U=G(X)ge u$ if and only if $Xge w$. Hence we have
$$begin{eqnarray}
P(Uge u)&=&P(Uge u, Xnotin B)+P(Uge u,Xin B)\
&=&P(Xge w, Xnotin B) +P(Xge w, Xin B)=P(Xge w)=1-G(w^-)=1-u.
end{eqnarray}$$
Secondly, assume $G{w}>0$ and $Uge u$. In case $Xnotin B$, $U=G(X)ge u$ is equivalent to $Xge w$. In case $Xin B$, we see that $X=b>w$ implies $Uge G(b^-)ge G(w)ge u$. A more subtle case is when $X=win B$. Then $U=G(w^-)+G{w}Vge u$ if and only if $Vge frac{u-G(w^-)}{G{w}}$. Summing this up, we have
$$begin{eqnarray}
P(Uge u)&=&P(Uge u, Xnotin B)+P(Uge u,Xin B)\
&=&P(Xge w, Xnotin B) +P(X> w, Xin B)+P(X= w, Vge frac{u-G(w^-)}{G{w}})\
&=&P(Xge w, Xnotin B) +P(X> w, Xin B)+P(X= w) left(1-frac{u-G(w^-)}{G{w}}right)\
&=&P(X>w) + G{w}left(1-frac{u-G(w^-)}{G{w}}right)\
&=&1-G(w)+G{w}-u+G(w^-)=1-u.
end{eqnarray}$$
Thus, it follows that $U$ is uniformly distributed on $[0,1]$.
answered Jan 18 at 20:03
SongSong
15.4k1736
edited Jan 18 at 20:15
answered Jan 18 at 20:03
SongSong
15.4k1736
answered Jan 18 at 20:03
SongSong
15.4k1736
answered Jan 18 at 20:03
SongSong
15.4k1736
15.4k1736
$begingroup$
Thanks a lot! Any hint on the quantile?
$endgroup$
– user17645
Jan 18 at 20:25
$begingroup$
The quantile function is as you defined. It has properties such as $$G^{-1}(a)=color{red} min{xinBbb R:G(x)ge a}$$ and $$G(x)ge a Longleftrightarrow xge G^{-1}(a).$$
$endgroup$
– Song
Jan 18 at 20:30
$begingroup$
I am having a very hard time using the properties you mentioned to prove that $X =G^{-1}(U)$ a.s.. Would you mind providing additional detailed explanation? Thanks!!!
$endgroup$
– user17645
Jan 22 at 8:23
$begingroup$
@user17645 I just realized that this was a part of your question ... Sorry for that. Now, to show $X=G^{-1}(U)$, first notice that $Ule G(X)$ with probability $1$. In view of the definition of $G^{-1}$, now it suffices to show that $$ x<X implies G(x)<U.$$ Analyze each case of $Xin B$ and $Xnotin B$. It would not be so difficult.
$endgroup$
– Song
Jan 22 at 16:38
add a comment |
$begingroup$
Thanks a lot! Any hint on the quantile?
$endgroup$
– user17645
Jan 18 at 20:25
$begingroup$
The quantile function is as you defined. It has properties such as $$G^{-1}(a)=color{red} min{xinBbb R:G(x)ge a}$$ and $$G(x)ge a Longleftrightarrow xge G^{-1}(a).$$
$endgroup$
– Song
Jan 18 at 20:30
$begingroup$
I am having a very hard time using the properties you mentioned to prove that $X =G^{-1}(U)$ a.s.. Would you mind providing additional detailed explanation? Thanks!!!
$endgroup$
– user17645
Jan 22 at 8:23
$begingroup$
@user17645 I just realized that this was a part of your question ... Sorry for that. Now, to show $X=G^{-1}(U)$, first notice that $Ule G(X)$ with probability $1$. In view of the definition of $G^{-1}$, now it suffices to show that $$ x<X implies G(x)<U.$$ Analyze each case of $Xin B$ and $Xnotin B$. It would not be so difficult.
$endgroup$
– Song
Jan 22 at 16:38
$begingroup$
Thanks a lot! Any hint on the quantile?
$endgroup$
– user17645
Jan 18 at 20:25
$begingroup$
Thanks a lot! Any hint on the quantile?
$endgroup$
– user17645
Jan 18 at 20:25
$begingroup$
The quantile function is as you defined. It has properties such as $$G^{-1}(a)=color{red} min{xinBbb R:G(x)ge a}$$ and $$G(x)ge a Longleftrightarrow xge G^{-1}(a).$$
$endgroup$
– Song
Jan 18 at 20:30
$begingroup$
The quantile function is as you defined. It has properties such as $$G^{-1}(a)=color{red} min{xinBbb R:G(x)ge a}$$ and $$G(x)ge a Longleftrightarrow xge G^{-1}(a).$$
$endgroup$
– Song
Jan 18 at 20:30
$begingroup$
I am having a very hard time using the properties you mentioned to prove that $X =G^{-1}(U)$ a.s.. Would you mind providing additional detailed explanation? Thanks!!!
$endgroup$
– user17645
Jan 22 at 8:23
$begingroup$
I am having a very hard time using the properties you mentioned to prove that $X =G^{-1}(U)$ a.s.. Would you mind providing additional detailed explanation? Thanks!!!
$endgroup$
– user17645
Jan 22 at 8:23
$begingroup$
@user17645 I just realized that this was a part of your question ... Sorry for that. Now, to show $X=G^{-1}(U)$, first notice that $Ule G(X)$ with probability $1$. In view of the definition of $G^{-1}$, now it suffices to show that $$ x<X implies G(x)<U.$$ Analyze each case of $Xin B$ and $Xnotin B$. It would not be so difficult.
$endgroup$
– Song
Jan 22 at 16:38
$begingroup$
@user17645 I just realized that this was a part of your question ... Sorry for that. Now, to show $X=G^{-1}(U)$, first notice that $Ule G(X)$ with probability $1$. In view of the definition of $G^{-1}$, now it suffices to show that $$ x<X implies G(x)<U.$$ Analyze each case of $Xin B$ and $Xnotin B$. It would not be so difficult.
$endgroup$
– Song
Jan 22 at 16:38
add a comment |
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