Distribution of $(XY)^Z$ if $(X,Y,Z)$ is i.i.d. uniform on $[0,1]$
$begingroup$
$X,Y$ and $Z$ are independent uniformly distributed on $[0,1]$
How is random variable $(XY)^Z$ distributed?
I had an idea to logarithm this and use convolution integral for the sum, but I'm not sure it's possible.
probability-theory probability-distributions
edited Jun 1 '17 at 21:08
Did
247k23223460
asked Dec 18 '12 at 23:31
XxxXxx
354110
$endgroup$
add a comment |
$begingroup$
$X,Y$ and $Z$ are independent uniformly distributed on $[0,1]$
How is random variable $(XY)^Z$ distributed?
I had an idea to logarithm this and use convolution integral for the sum, but I'm not sure it's possible.
probability-theory probability-distributions
edited Jun 1 '17 at 21:08
Did
247k23223460
asked Dec 18 '12 at 23:31
XxxXxx
354110
$endgroup$
$begingroup$
Why do you need to know? Also, does $(XY)^{frac{1}{2}}$ mean $sqrt{XY}$ or $-sqrt{XY}$.
$endgroup$
– Dilip Sarwate
Dec 18 '12 at 23:45
add a comment |
$begingroup$
$X,Y$ and $Z$ are independent uniformly distributed on $[0,1]$
How is random variable $(XY)^Z$ distributed?
I had an idea to logarithm this and use convolution integral for the sum, but I'm not sure it's possible.
probability-theory probability-distributions
edited Jun 1 '17 at 21:08
Did
247k23223460
asked Dec 18 '12 at 23:31
XxxXxx
354110
$endgroup$
$X,Y$ and $Z$ are independent uniformly distributed on $[0,1]$
How is random variable $(XY)^Z$ distributed?
I had an idea to logarithm this and use convolution integral for the sum, but I'm not sure it's possible.
probability-theory probability-distributions
probability-theory probability-distributions
edited Jun 1 '17 at 21:08
Did
247k23223460
asked Dec 18 '12 at 23:31
XxxXxx
354110
edited Jun 1 '17 at 21:08
Did
247k23223460
asked Dec 18 '12 at 23:31
XxxXxx
354110
edited Jun 1 '17 at 21:08
Did
247k23223460
edited Jun 1 '17 at 21:08
Did
247k23223460
edited Jun 1 '17 at 21:08
Did
247k23223460
247k23223460
asked Dec 18 '12 at 23:31
XxxXxx
354110
asked Dec 18 '12 at 23:31
XxxXxx
354110
asked Dec 18 '12 at 23:31
XxxXxx
354110
354110
$begingroup$
Why do you need to know? Also, does $(XY)^{frac{1}{2}}$ mean $sqrt{XY}$ or $-sqrt{XY}$.
$endgroup$
– Dilip Sarwate
Dec 18 '12 at 23:45
add a comment |
$begingroup$
Why do you need to know? Also, does $(XY)^{frac{1}{2}}$ mean $sqrt{XY}$ or $-sqrt{XY}$.
$endgroup$
– Dilip Sarwate
Dec 18 '12 at 23:45
$begingroup$
Why do you need to know? Also, does $(XY)^{frac{1}{2}}$ mean $sqrt{XY}$ or $-sqrt{XY}$.
$endgroup$
– Dilip Sarwate
Dec 18 '12 at 23:45
$begingroup$
Why do you need to know? Also, does $(XY)^{frac{1}{2}}$ mean $sqrt{XY}$ or $-sqrt{XY}$.
$endgroup$
– Dilip Sarwate
Dec 18 '12 at 23:45
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Hints:
The random variable $X$ is uniform on $(0,1)$ if and only if $-log X$ is exponential with parameter $1$.
If $U$ and $V$ are independent and exponential with parameter $1$, then $U+V$ is gamma distributed $(2,1)$, that is, with density $wmapsto wmathrm e^{-w}mathbf 1_{wgt0}$.
If $W$ is gamma distributed $(2,1)$ and $T$ is uniform on $(0,1)$ and independent of $W$, then $WU$ is exponential with parameter $1$.
Conclusion:
- If $X$, $Y$ and $Z$ are independent and uniform on $(0,1)$, then $(XY)^Z$ is uniform on $(0,1)$.
answered Dec 19 '12 at 0:01
DidDid
247k23223460
$endgroup$
$begingroup$
+1: this is a nicer argument than my calculation... Even though they rely on fact which the OP might or might not have already proven.
$endgroup$
– Fabian
Dec 19 '12 at 0:02
add a comment |
$begingroup$
Given the simplicity of the result there must be a nice short way to obtain it. However, I did not find one so I present the long and complicated calculation.
The distribution of the random variable $W=(XY)^Z$ is given by:
$$begin{align}P(wgeq W) &= int_0^1!dxint_0^1!dyint_0^1!dz, theta(w-(xy)^z)\
&= int_0^1!dxint_0^1!dy max{1-log_{xy} w,0}\
&=int_0^1!detaint_eta^1!frac{dx}{x}max{1-log_{eta} w,0} \
&=-int_0^w!deta log eta (1-log_{eta} w)\
&=w.
end{align}$$
with $eta=xy$.
Thus the variable $W$ is also uniformly distributed (between 0 and 1).
answered Dec 19 '12 at 0:00
FabianFabian
19.7k3674
$endgroup$
add a comment |
$begingroup$
Using the definition of weak convergence, it is so easy. First, for any positive integer $kge 0$ we have $E{W^k}=1/(k+1)=EU^k$. Hence, for any polynomial $f(x)$, we have $Ef(W)=Ef(U)$. For any bounded and continuous function $g(cdot)$, we can find a polynomial function $f(cdot)$ such that $f$ can approximate $g$ uniformly by Weierstrass's theorem. Thus, $Eg(W)=Eg(U)$. So $Wsim U(0,1)$.
edited Jan 21 at 19:03
Mars Plastic
865
answered Apr 18 '13 at 13:57
user73223user73223
1
$endgroup$
$begingroup$
You can format your answer using latex.
$endgroup$
– user60610
Apr 18 '13 at 14:17
$begingroup$
Hmmm... By the way, is this so obvious that E{W^k}=1/(k+1)?
$endgroup$
– Did
Nov 27 '18 at 19:00
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hints:
The random variable $X$ is uniform on $(0,1)$ if and only if $-log X$ is exponential with parameter $1$.
If $U$ and $V$ are independent and exponential with parameter $1$, then $U+V$ is gamma distributed $(2,1)$, that is, with density $wmapsto wmathrm e^{-w}mathbf 1_{wgt0}$.
If $W$ is gamma distributed $(2,1)$ and $T$ is uniform on $(0,1)$ and independent of $W$, then $WU$ is exponential with parameter $1$.
Conclusion:
- If $X$, $Y$ and $Z$ are independent and uniform on $(0,1)$, then $(XY)^Z$ is uniform on $(0,1)$.
answered Dec 19 '12 at 0:01
DidDid
247k23223460
$endgroup$
$begingroup$
+1: this is a nicer argument than my calculation... Even though they rely on fact which the OP might or might not have already proven.
$endgroup$
– Fabian
Dec 19 '12 at 0:02
add a comment |
$begingroup$
Hints:
The random variable $X$ is uniform on $(0,1)$ if and only if $-log X$ is exponential with parameter $1$.
If $U$ and $V$ are independent and exponential with parameter $1$, then $U+V$ is gamma distributed $(2,1)$, that is, with density $wmapsto wmathrm e^{-w}mathbf 1_{wgt0}$.
If $W$ is gamma distributed $(2,1)$ and $T$ is uniform on $(0,1)$ and independent of $W$, then $WU$ is exponential with parameter $1$.
Conclusion:
- If $X$, $Y$ and $Z$ are independent and uniform on $(0,1)$, then $(XY)^Z$ is uniform on $(0,1)$.
answered Dec 19 '12 at 0:01
DidDid
247k23223460
$endgroup$
$begingroup$
+1: this is a nicer argument than my calculation... Even though they rely on fact which the OP might or might not have already proven.
$endgroup$
– Fabian
Dec 19 '12 at 0:02
add a comment |
$begingroup$
Hints:
The random variable $X$ is uniform on $(0,1)$ if and only if $-log X$ is exponential with parameter $1$.
If $U$ and $V$ are independent and exponential with parameter $1$, then $U+V$ is gamma distributed $(2,1)$, that is, with density $wmapsto wmathrm e^{-w}mathbf 1_{wgt0}$.
If $W$ is gamma distributed $(2,1)$ and $T$ is uniform on $(0,1)$ and independent of $W$, then $WU$ is exponential with parameter $1$.
Conclusion:
- If $X$, $Y$ and $Z$ are independent and uniform on $(0,1)$, then $(XY)^Z$ is uniform on $(0,1)$.
answered Dec 19 '12 at 0:01
DidDid
247k23223460
$endgroup$
Hints:
The random variable $X$ is uniform on $(0,1)$ if and only if $-log X$ is exponential with parameter $1$.
If $U$ and $V$ are independent and exponential with parameter $1$, then $U+V$ is gamma distributed $(2,1)$, that is, with density $wmapsto wmathrm e^{-w}mathbf 1_{wgt0}$.
If $W$ is gamma distributed $(2,1)$ and $T$ is uniform on $(0,1)$ and independent of $W$, then $WU$ is exponential with parameter $1$.
Conclusion:
- If $X$, $Y$ and $Z$ are independent and uniform on $(0,1)$, then $(XY)^Z$ is uniform on $(0,1)$.
answered Dec 19 '12 at 0:01
DidDid
247k23223460
answered Dec 19 '12 at 0:01
DidDid
247k23223460
answered Dec 19 '12 at 0:01
DidDid
247k23223460
answered Dec 19 '12 at 0:01
DidDid
247k23223460
247k23223460
$begingroup$
+1: this is a nicer argument than my calculation... Even though they rely on fact which the OP might or might not have already proven.
$endgroup$
– Fabian
Dec 19 '12 at 0:02
add a comment |
$begingroup$
+1: this is a nicer argument than my calculation... Even though they rely on fact which the OP might or might not have already proven.
$endgroup$
– Fabian
Dec 19 '12 at 0:02
$begingroup$
+1: this is a nicer argument than my calculation... Even though they rely on fact which the OP might or might not have already proven.
$endgroup$
– Fabian
Dec 19 '12 at 0:02
$begingroup$
+1: this is a nicer argument than my calculation... Even though they rely on fact which the OP might or might not have already proven.
$endgroup$
– Fabian
Dec 19 '12 at 0:02
add a comment |
$begingroup$
Given the simplicity of the result there must be a nice short way to obtain it. However, I did not find one so I present the long and complicated calculation.
The distribution of the random variable $W=(XY)^Z$ is given by:
$$begin{align}P(wgeq W) &= int_0^1!dxint_0^1!dyint_0^1!dz, theta(w-(xy)^z)\
&= int_0^1!dxint_0^1!dy max{1-log_{xy} w,0}\
&=int_0^1!detaint_eta^1!frac{dx}{x}max{1-log_{eta} w,0} \
&=-int_0^w!deta log eta (1-log_{eta} w)\
&=w.
end{align}$$
with $eta=xy$.
Thus the variable $W$ is also uniformly distributed (between 0 and 1).
answered Dec 19 '12 at 0:00
FabianFabian
19.7k3674
$endgroup$
add a comment |
$begingroup$
Given the simplicity of the result there must be a nice short way to obtain it. However, I did not find one so I present the long and complicated calculation.
The distribution of the random variable $W=(XY)^Z$ is given by:
$$begin{align}P(wgeq W) &= int_0^1!dxint_0^1!dyint_0^1!dz, theta(w-(xy)^z)\
&= int_0^1!dxint_0^1!dy max{1-log_{xy} w,0}\
&=int_0^1!detaint_eta^1!frac{dx}{x}max{1-log_{eta} w,0} \
&=-int_0^w!deta log eta (1-log_{eta} w)\
&=w.
end{align}$$
with $eta=xy$.
Thus the variable $W$ is also uniformly distributed (between 0 and 1).
answered Dec 19 '12 at 0:00
FabianFabian
19.7k3674
$endgroup$
add a comment |
$begingroup$
Given the simplicity of the result there must be a nice short way to obtain it. However, I did not find one so I present the long and complicated calculation.
The distribution of the random variable $W=(XY)^Z$ is given by:
$$begin{align}P(wgeq W) &= int_0^1!dxint_0^1!dyint_0^1!dz, theta(w-(xy)^z)\
&= int_0^1!dxint_0^1!dy max{1-log_{xy} w,0}\
&=int_0^1!detaint_eta^1!frac{dx}{x}max{1-log_{eta} w,0} \
&=-int_0^w!deta log eta (1-log_{eta} w)\
&=w.
end{align}$$
with $eta=xy$.
Thus the variable $W$ is also uniformly distributed (between 0 and 1).
answered Dec 19 '12 at 0:00
FabianFabian
19.7k3674
$endgroup$
Given the simplicity of the result there must be a nice short way to obtain it. However, I did not find one so I present the long and complicated calculation.
The distribution of the random variable $W=(XY)^Z$ is given by:
$$begin{align}P(wgeq W) &= int_0^1!dxint_0^1!dyint_0^1!dz, theta(w-(xy)^z)\
&= int_0^1!dxint_0^1!dy max{1-log_{xy} w,0}\
&=int_0^1!detaint_eta^1!frac{dx}{x}max{1-log_{eta} w,0} \
&=-int_0^w!deta log eta (1-log_{eta} w)\
&=w.
end{align}$$
with $eta=xy$.
Thus the variable $W$ is also uniformly distributed (between 0 and 1).
answered Dec 19 '12 at 0:00
FabianFabian
19.7k3674
answered Dec 19 '12 at 0:00
FabianFabian
19.7k3674
answered Dec 19 '12 at 0:00
FabianFabian
19.7k3674
answered Dec 19 '12 at 0:00
FabianFabian
19.7k3674
19.7k3674
add a comment |
add a comment |
$begingroup$
Using the definition of weak convergence, it is so easy. First, for any positive integer $kge 0$ we have $E{W^k}=1/(k+1)=EU^k$. Hence, for any polynomial $f(x)$, we have $Ef(W)=Ef(U)$. For any bounded and continuous function $g(cdot)$, we can find a polynomial function $f(cdot)$ such that $f$ can approximate $g$ uniformly by Weierstrass's theorem. Thus, $Eg(W)=Eg(U)$. So $Wsim U(0,1)$.
edited Jan 21 at 19:03
Mars Plastic
865
answered Apr 18 '13 at 13:57
user73223user73223
1
$endgroup$
$begingroup$
You can format your answer using latex.
$endgroup$
– user60610
Apr 18 '13 at 14:17
$begingroup$
Hmmm... By the way, is this so obvious that E{W^k}=1/(k+1)?
$endgroup$
– Did
Nov 27 '18 at 19:00
add a comment |
$begingroup$
Using the definition of weak convergence, it is so easy. First, for any positive integer $kge 0$ we have $E{W^k}=1/(k+1)=EU^k$. Hence, for any polynomial $f(x)$, we have $Ef(W)=Ef(U)$. For any bounded and continuous function $g(cdot)$, we can find a polynomial function $f(cdot)$ such that $f$ can approximate $g$ uniformly by Weierstrass's theorem. Thus, $Eg(W)=Eg(U)$. So $Wsim U(0,1)$.
edited Jan 21 at 19:03
Mars Plastic
865
answered Apr 18 '13 at 13:57
user73223user73223
1
$endgroup$
$begingroup$
You can format your answer using latex.
$endgroup$
– user60610
Apr 18 '13 at 14:17
$begingroup$
Hmmm... By the way, is this so obvious that E{W^k}=1/(k+1)?
$endgroup$
– Did
Nov 27 '18 at 19:00
add a comment |
$begingroup$
Using the definition of weak convergence, it is so easy. First, for any positive integer $kge 0$ we have $E{W^k}=1/(k+1)=EU^k$. Hence, for any polynomial $f(x)$, we have $Ef(W)=Ef(U)$. For any bounded and continuous function $g(cdot)$, we can find a polynomial function $f(cdot)$ such that $f$ can approximate $g$ uniformly by Weierstrass's theorem. Thus, $Eg(W)=Eg(U)$. So $Wsim U(0,1)$.
edited Jan 21 at 19:03
Mars Plastic
865
answered Apr 18 '13 at 13:57
user73223user73223
1
$endgroup$
Using the definition of weak convergence, it is so easy. First, for any positive integer $kge 0$ we have $E{W^k}=1/(k+1)=EU^k$. Hence, for any polynomial $f(x)$, we have $Ef(W)=Ef(U)$. For any bounded and continuous function $g(cdot)$, we can find a polynomial function $f(cdot)$ such that $f$ can approximate $g$ uniformly by Weierstrass's theorem. Thus, $Eg(W)=Eg(U)$. So $Wsim U(0,1)$.
edited Jan 21 at 19:03
Mars Plastic
865
answered Apr 18 '13 at 13:57
user73223user73223
1
edited Jan 21 at 19:03
Mars Plastic
865
edited Jan 21 at 19:03
Mars Plastic
865
edited Jan 21 at 19:03
Mars Plastic
865
865
answered Apr 18 '13 at 13:57
user73223user73223
1
answered Apr 18 '13 at 13:57
user73223user73223
1
answered Apr 18 '13 at 13:57
user73223user73223
1
1
$begingroup$
You can format your answer using latex.
$endgroup$
– user60610
Apr 18 '13 at 14:17
$begingroup$
Hmmm... By the way, is this so obvious that E{W^k}=1/(k+1)?
$endgroup$
– Did
Nov 27 '18 at 19:00
add a comment |
$begingroup$
You can format your answer using latex.
$endgroup$
– user60610
Apr 18 '13 at 14:17
$begingroup$
Hmmm... By the way, is this so obvious that E{W^k}=1/(k+1)?
$endgroup$
– Did
Nov 27 '18 at 19:00
$begingroup$
You can format your answer using latex.
$endgroup$
– user60610
Apr 18 '13 at 14:17
$begingroup$
You can format your answer using latex.
$endgroup$
– user60610
Apr 18 '13 at 14:17
$begingroup$
Hmmm... By the way, is this so obvious that E{W^k}=1/(k+1)?
$endgroup$
– Did
Nov 27 '18 at 19:00
$begingroup$
Hmmm... By the way, is this so obvious that E{W^k}=1/(k+1)?
$endgroup$
– Did
Nov 27 '18 at 19:00
add a comment |
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$begingroup$
Why do you need to know? Also, does $(XY)^{frac{1}{2}}$ mean $sqrt{XY}$ or $-sqrt{XY}$.
$endgroup$
– Dilip Sarwate
Dec 18 '12 at 23:45