Transformation of Random variable $Y=-2ln(F(x))$ [on hold]
Let $X$ is a continuous Random variable. with strictly increasing function cumulative distribution function $F(x)$.
Find and recognise the distribution of random variable $Y=-2ln(F(x))$.
I need some help here please.
probability transformation
asked Jan 4 at 0:52
kotsos sgouras
35
put on hold as off-topic by Did, StubbornAtom, Paul Frost, Pierre-Guy Plamondon, Cesareo Jan 4 at 12:50
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, StubbornAtom, Paul Frost, Pierre-Guy Plamondon, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
Let $X$ is a continuous Random variable. with strictly increasing function cumulative distribution function $F(x)$.
Find and recognise the distribution of random variable $Y=-2ln(F(x))$.
I need some help here please.
probability transformation
asked Jan 4 at 0:52
kotsos sgouras
35
put on hold as off-topic by Did, StubbornAtom, Paul Frost, Pierre-Guy Plamondon, Cesareo Jan 4 at 12:50
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, StubbornAtom, Paul Frost, Pierre-Guy Plamondon, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
Can we assume $F$ is continuous?
– kimchi lover
Jan 4 at 1:01
Yes X is continuous
– kotsos sgouras
Jan 4 at 7:22
add a comment |
Let $X$ is a continuous Random variable. with strictly increasing function cumulative distribution function $F(x)$.
Find and recognise the distribution of random variable $Y=-2ln(F(x))$.
I need some help here please.
probability transformation
asked Jan 4 at 0:52
kotsos sgouras
35
Let $X$ is a continuous Random variable. with strictly increasing function cumulative distribution function $F(x)$.
Find and recognise the distribution of random variable $Y=-2ln(F(x))$.
I need some help here please.
probability transformation
probability transformation
asked Jan 4 at 0:52
kotsos sgouras
35
asked Jan 4 at 0:52
kotsos sgouras
35
edited Jan 4 at 7:28
asked Jan 4 at 0:52
kotsos sgouras
35
asked Jan 4 at 0:52
kotsos sgouras
35
asked Jan 4 at 0:52
kotsos sgouras
35
35
put on hold as off-topic by Did, StubbornAtom, Paul Frost, Pierre-Guy Plamondon, Cesareo Jan 4 at 12:50
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, StubbornAtom, Paul Frost, Pierre-Guy Plamondon, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as off-topic by Did, StubbornAtom, Paul Frost, Pierre-Guy Plamondon, Cesareo Jan 4 at 12:50
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, StubbornAtom, Paul Frost, Pierre-Guy Plamondon, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
Can we assume $F$ is continuous?
– kimchi lover
Jan 4 at 1:01
Yes X is continuous
– kotsos sgouras
Jan 4 at 7:22
add a comment |
Can we assume $F$ is continuous?
– kimchi lover
Jan 4 at 1:01
Yes X is continuous
– kotsos sgouras
Jan 4 at 7:22
Can we assume $F$ is continuous?
– kimchi lover
Jan 4 at 1:01
Can we assume $F$ is continuous?
– kimchi lover
Jan 4 at 1:01
Yes X is continuous
– kotsos sgouras
Jan 4 at 7:22
Yes X is continuous
– kotsos sgouras
Jan 4 at 7:22
add a comment |
1 Answer
1
active
oldest
votes
It's easy if $F$ is continuous. In that case $P(F(X)leq t)=P(Xleq F^{-1}(t))=F(F^{-1}(t))=t$ for $0<t<1.$ So, $F(X)sim U(0,1).$ $P(-2log F(X)leq y/2)=P(log F(X)geq -y/2)=P(F(X)geq e^{-y/2})=1-e^{-y/2}$ for $y>0.$ So, $Ysim Exp(2)$ or $Ysim chi^2_2.$
But if $F$ is not continuous, i.e, there is jump in $F$, then $F(X)$ might not be uniformly distributed.
answered Jan 4 at 1:17
John_Wick
1,436111
We can say $P{Y leq y}=1-F(G(e^{-y/2)}))$ whether or not $F$ is continuous, where $G(t)=inf {x:F(x) geq t}$. If $F$ is continuous then $Fcirc G$ is the identity function.
– Kavi Rama Murthy
Jan 4 at 5:56
@kotsossgouras The first $y/2$ is a misprint; it should say $y$.
– J.G.
Jan 4 at 7:55
how we can show that F^-1(F(X))=X ??
– kotsos sgouras
Jan 4 at 12:34
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
It's easy if $F$ is continuous. In that case $P(F(X)leq t)=P(Xleq F^{-1}(t))=F(F^{-1}(t))=t$ for $0<t<1.$ So, $F(X)sim U(0,1).$ $P(-2log F(X)leq y/2)=P(log F(X)geq -y/2)=P(F(X)geq e^{-y/2})=1-e^{-y/2}$ for $y>0.$ So, $Ysim Exp(2)$ or $Ysim chi^2_2.$
But if $F$ is not continuous, i.e, there is jump in $F$, then $F(X)$ might not be uniformly distributed.
answered Jan 4 at 1:17
John_Wick
1,436111
We can say $P{Y leq y}=1-F(G(e^{-y/2)}))$ whether or not $F$ is continuous, where $G(t)=inf {x:F(x) geq t}$. If $F$ is continuous then $Fcirc G$ is the identity function.
– Kavi Rama Murthy
Jan 4 at 5:56
@kotsossgouras The first $y/2$ is a misprint; it should say $y$.
– J.G.
Jan 4 at 7:55
how we can show that F^-1(F(X))=X ??
– kotsos sgouras
Jan 4 at 12:34
add a comment |
It's easy if $F$ is continuous. In that case $P(F(X)leq t)=P(Xleq F^{-1}(t))=F(F^{-1}(t))=t$ for $0<t<1.$ So, $F(X)sim U(0,1).$ $P(-2log F(X)leq y/2)=P(log F(X)geq -y/2)=P(F(X)geq e^{-y/2})=1-e^{-y/2}$ for $y>0.$ So, $Ysim Exp(2)$ or $Ysim chi^2_2.$
But if $F$ is not continuous, i.e, there is jump in $F$, then $F(X)$ might not be uniformly distributed.
answered Jan 4 at 1:17
John_Wick
1,436111
We can say $P{Y leq y}=1-F(G(e^{-y/2)}))$ whether or not $F$ is continuous, where $G(t)=inf {x:F(x) geq t}$. If $F$ is continuous then $Fcirc G$ is the identity function.
– Kavi Rama Murthy
Jan 4 at 5:56
@kotsossgouras The first $y/2$ is a misprint; it should say $y$.
– J.G.
Jan 4 at 7:55
how we can show that F^-1(F(X))=X ??
– kotsos sgouras
Jan 4 at 12:34
add a comment |
It's easy if $F$ is continuous. In that case $P(F(X)leq t)=P(Xleq F^{-1}(t))=F(F^{-1}(t))=t$ for $0<t<1.$ So, $F(X)sim U(0,1).$ $P(-2log F(X)leq y/2)=P(log F(X)geq -y/2)=P(F(X)geq e^{-y/2})=1-e^{-y/2}$ for $y>0.$ So, $Ysim Exp(2)$ or $Ysim chi^2_2.$
But if $F$ is not continuous, i.e, there is jump in $F$, then $F(X)$ might not be uniformly distributed.
answered Jan 4 at 1:17
John_Wick
1,436111
It's easy if $F$ is continuous. In that case $P(F(X)leq t)=P(Xleq F^{-1}(t))=F(F^{-1}(t))=t$ for $0<t<1.$ So, $F(X)sim U(0,1).$ $P(-2log F(X)leq y/2)=P(log F(X)geq -y/2)=P(F(X)geq e^{-y/2})=1-e^{-y/2}$ for $y>0.$ So, $Ysim Exp(2)$ or $Ysim chi^2_2.$
But if $F$ is not continuous, i.e, there is jump in $F$, then $F(X)$ might not be uniformly distributed.
answered Jan 4 at 1:17
John_Wick
1,436111
answered Jan 4 at 1:17
John_Wick
1,436111
answered Jan 4 at 1:17
John_Wick
1,436111
answered Jan 4 at 1:17
John_Wick
1,436111
1,436111
We can say $P{Y leq y}=1-F(G(e^{-y/2)}))$ whether or not $F$ is continuous, where $G(t)=inf {x:F(x) geq t}$. If $F$ is continuous then $Fcirc G$ is the identity function.
– Kavi Rama Murthy
Jan 4 at 5:56
@kotsossgouras The first $y/2$ is a misprint; it should say $y$.
– J.G.
Jan 4 at 7:55
how we can show that F^-1(F(X))=X ??
– kotsos sgouras
Jan 4 at 12:34
add a comment |
We can say $P{Y leq y}=1-F(G(e^{-y/2)}))$ whether or not $F$ is continuous, where $G(t)=inf {x:F(x) geq t}$. If $F$ is continuous then $Fcirc G$ is the identity function.
– Kavi Rama Murthy
Jan 4 at 5:56
@kotsossgouras The first $y/2$ is a misprint; it should say $y$.
– J.G.
Jan 4 at 7:55
how we can show that F^-1(F(X))=X ??
– kotsos sgouras
Jan 4 at 12:34
We can say $P{Y leq y}=1-F(G(e^{-y/2)}))$ whether or not $F$ is continuous, where $G(t)=inf {x:F(x) geq t}$. If $F$ is continuous then $Fcirc G$ is the identity function.
– Kavi Rama Murthy
Jan 4 at 5:56
We can say $P{Y leq y}=1-F(G(e^{-y/2)}))$ whether or not $F$ is continuous, where $G(t)=inf {x:F(x) geq t}$. If $F$ is continuous then $Fcirc G$ is the identity function.
– Kavi Rama Murthy
Jan 4 at 5:56
@kotsossgouras The first $y/2$ is a misprint; it should say $y$.
– J.G.
Jan 4 at 7:55
@kotsossgouras The first $y/2$ is a misprint; it should say $y$.
– J.G.
Jan 4 at 7:55
how we can show that F^-1(F(X))=X ??
– kotsos sgouras
Jan 4 at 12:34
how we can show that F^-1(F(X))=X ??
– kotsos sgouras
Jan 4 at 12:34
add a comment |
Can we assume $F$ is continuous?
– kimchi lover
Jan 4 at 1:01
Yes X is continuous
– kotsos sgouras
Jan 4 at 7:22