Mixing
What is the distribution followed by this function of random variables
$begingroup$
$X$ and $Y$ are independently distributed $U(0,1)$ random variables (Uniform distribution)
Find the Variance of the random variable
$U = dfrac{ln(X)}{( ln(X)+ln(1-Y) )}$
I know $-ln(X)$ follows $exp(1)$ and $(1-Y)$ is also $U(0,1)$
Therefore it becomes $exp(1)$ in the numerator and $text{gamma}(1,2)$ in the denominator.
I dont know what to do next.
probability statistics probability-distributions random-variables statistical-inference
edited Feb 3 at 3:35
Thomas Shelby
4,7382727
asked Feb 2 at 18:00
Vishesh ChughVishesh Chugh
65
$endgroup$
|
show 4 more comments
$begingroup$
$X$ and $Y$ are independently distributed $U(0,1)$ random variables (Uniform distribution)
Find the Variance of the random variable
$U = dfrac{ln(X)}{( ln(X)+ln(1-Y) )}$
I know $-ln(X)$ follows $exp(1)$ and $(1-Y)$ is also $U(0,1)$
Therefore it becomes $exp(1)$ in the numerator and $text{gamma}(1,2)$ in the denominator.
I dont know what to do next.
probability statistics probability-distributions random-variables statistical-inference
edited Feb 3 at 3:35
Thomas Shelby
4,7382727
asked Feb 2 at 18:00
Vishesh ChughVishesh Chugh
65
$endgroup$
1
$begingroup$
You wrote "and $ln(1-Y)$ is also $U(0,1)$. But this is not true. Write your ratio as $S/(S+T)$ where $S,T$ are iid exponential rvs.
$endgroup$
– kimchi lover
Feb 2 at 18:08
1
$begingroup$
@kimchilover - presumably "and also $-ln(1-Y) sim exp(1)$" or something similar was intended
$endgroup$
– Henry
Feb 2 at 18:13
$begingroup$
@Henry I agree.
$endgroup$
– kimchi lover
Feb 2 at 18:15
1
$begingroup$
You will see $U$ has a standard distribution if you derive it (say, by a change of variables or using well-known relations between distributions), from which the variance follows immediately.
$endgroup$
– StubbornAtom
Feb 2 at 18:34
1
$begingroup$
@Vishesh You are not showing your work/attempts, so it's hard to know where you are stuck.
$endgroup$
– StubbornAtom
Feb 2 at 18:39
|
show 4 more comments
$begingroup$
$X$ and $Y$ are independently distributed $U(0,1)$ random variables (Uniform distribution)
Find the Variance of the random variable
$U = dfrac{ln(X)}{( ln(X)+ln(1-Y) )}$
I know $-ln(X)$ follows $exp(1)$ and $(1-Y)$ is also $U(0,1)$
Therefore it becomes $exp(1)$ in the numerator and $text{gamma}(1,2)$ in the denominator.
I dont know what to do next.
probability statistics probability-distributions random-variables statistical-inference
edited Feb 3 at 3:35
Thomas Shelby
4,7382727
asked Feb 2 at 18:00
Vishesh ChughVishesh Chugh
65
$endgroup$
$X$ and $Y$ are independently distributed $U(0,1)$ random variables (Uniform distribution)
Find the Variance of the random variable
$U = dfrac{ln(X)}{( ln(X)+ln(1-Y) )}$
I know $-ln(X)$ follows $exp(1)$ and $(1-Y)$ is also $U(0,1)$
Therefore it becomes $exp(1)$ in the numerator and $text{gamma}(1,2)$ in the denominator.
I dont know what to do next.
probability statistics probability-distributions random-variables statistical-inference
probability statistics probability-distributions random-variables statistical-inference
edited Feb 3 at 3:35
Thomas Shelby
4,7382727
asked Feb 2 at 18:00
Vishesh ChughVishesh Chugh
65
edited Feb 3 at 3:35
Thomas Shelby
4,7382727
asked Feb 2 at 18:00
Vishesh ChughVishesh Chugh
65
edited Feb 3 at 3:35
Thomas Shelby
4,7382727
edited Feb 3 at 3:35
Thomas Shelby
4,7382727
edited Feb 3 at 3:35
Thomas Shelby
4,7382727
4,7382727
asked Feb 2 at 18:00
Vishesh ChughVishesh Chugh
65
asked Feb 2 at 18:00
Vishesh ChughVishesh Chugh
65
asked Feb 2 at 18:00
Vishesh ChughVishesh Chugh
65
65
1
$begingroup$
You wrote "and $ln(1-Y)$ is also $U(0,1)$. But this is not true. Write your ratio as $S/(S+T)$ where $S,T$ are iid exponential rvs.
$endgroup$
– kimchi lover
Feb 2 at 18:08
1
$begingroup$
@kimchilover - presumably "and also $-ln(1-Y) sim exp(1)$" or something similar was intended
$endgroup$
– Henry
Feb 2 at 18:13
$begingroup$
@Henry I agree.
$endgroup$
– kimchi lover
Feb 2 at 18:15
1
$begingroup$
You will see $U$ has a standard distribution if you derive it (say, by a change of variables or using well-known relations between distributions), from which the variance follows immediately.
$endgroup$
– StubbornAtom
Feb 2 at 18:34
1
$begingroup$
@Vishesh You are not showing your work/attempts, so it's hard to know where you are stuck.
$endgroup$
– StubbornAtom
Feb 2 at 18:39
|
show 4 more comments
1
$begingroup$
You wrote "and $ln(1-Y)$ is also $U(0,1)$. But this is not true. Write your ratio as $S/(S+T)$ where $S,T$ are iid exponential rvs.
$endgroup$
– kimchi lover
Feb 2 at 18:08
1
$begingroup$
@kimchilover - presumably "and also $-ln(1-Y) sim exp(1)$" or something similar was intended
$endgroup$
– Henry
Feb 2 at 18:13
$begingroup$
@Henry I agree.
$endgroup$
– kimchi lover
Feb 2 at 18:15
1
$begingroup$
You will see $U$ has a standard distribution if you derive it (say, by a change of variables or using well-known relations between distributions), from which the variance follows immediately.
$endgroup$
– StubbornAtom
Feb 2 at 18:34
1
$begingroup$
@Vishesh You are not showing your work/attempts, so it's hard to know where you are stuck.
$endgroup$
– StubbornAtom
Feb 2 at 18:39
1
1
$begingroup$
You wrote "and $ln(1-Y)$ is also $U(0,1)$. But this is not true. Write your ratio as $S/(S+T)$ where $S,T$ are iid exponential rvs.
$endgroup$
– kimchi lover
Feb 2 at 18:08
$begingroup$
You wrote "and $ln(1-Y)$ is also $U(0,1)$. But this is not true. Write your ratio as $S/(S+T)$ where $S,T$ are iid exponential rvs.
$endgroup$
– kimchi lover
Feb 2 at 18:08
1
1
$begingroup$
@kimchilover - presumably "and also $-ln(1-Y) sim exp(1)$" or something similar was intended
$endgroup$
– Henry
Feb 2 at 18:13
$begingroup$
@kimchilover - presumably "and also $-ln(1-Y) sim exp(1)$" or something similar was intended
$endgroup$
– Henry
Feb 2 at 18:13
$begingroup$
@Henry I agree.
$endgroup$
– kimchi lover
Feb 2 at 18:15
$begingroup$
@Henry I agree.
$endgroup$
– kimchi lover
Feb 2 at 18:15
1
1
$begingroup$
You will see $U$ has a standard distribution if you derive it (say, by a change of variables or using well-known relations between distributions), from which the variance follows immediately.
$endgroup$
– StubbornAtom
Feb 2 at 18:34
$begingroup$
You will see $U$ has a standard distribution if you derive it (say, by a change of variables or using well-known relations between distributions), from which the variance follows immediately.
$endgroup$
– StubbornAtom
Feb 2 at 18:34
1
1
$begingroup$
@Vishesh You are not showing your work/attempts, so it's hard to know where you are stuck.
$endgroup$
– StubbornAtom
Feb 2 at 18:39
$begingroup$
@Vishesh You are not showing your work/attempts, so it's hard to know where you are stuck.
$endgroup$
– StubbornAtom
Feb 2 at 18:39
|
show 4 more comments
2 Answers
2
active
oldest
votes
$begingroup$
Write as
$$
U=frac{-ln X}{-ln X-ln (1-Y)}
$$
As you noted $-ln Xsim exp(1)$ and because $1-Ysim text{Unif}(0,1)$ it folows that $-ln(1-Y)sim exp(1)$. Since $-ln X$ and $-ln (1-Y)$ are independent it follows that $Usim text{Beta}(1,1)$ which is the same as a uniform random variable on $(0,1)$.
answered Feb 2 at 18:49
Foobaz JohnFoobaz John
22.9k41552
$endgroup$
$begingroup$
I was stuck on this for so long. Thanks a lot mate.
$endgroup$
– Vishesh Chugh
Feb 3 at 2:09
2
$begingroup$
@VisheshChugh Sorry but if you were stuck before, you should still be stuck since this answer does not show you how to prove the result but merely invokes a more general fact (which you probably do not know how to prove either) to deduce this one. Strange world...
$endgroup$
– Did
Feb 3 at 8:57
$begingroup$
@Did There's a thing known as confusion and that shouldn't be so hard for you to understand as you so much already. Don't know why you'd be so irrelevant. Strange world. :) :)
$endgroup$
– Vishesh Chugh
Feb 14 at 9:57
add a comment |
$begingroup$
$G(lambda)=e^{-x}x^{lambda-1};x>0,lambda>0 $ and $B_1(u,v)={x^{u-1}}(1-x)^{v-1} ; 0<x<1 ,u>0,v>0 $
$X sim G(lambda) $ and $Ysim G(m)$
$dfrac{X}{X+Y} sim B_1(lambda,m)$
And good luck for JAM $2019$
answered Feb 3 at 3:41
Daman deepDaman deep
756420
$endgroup$
$begingroup$
Thanks mate. Haha how do you know? You taking it too? This year?
$endgroup$
– Vishesh Chugh
Feb 3 at 10:51
$begingroup$
This question came in any previous year, I solved it few days back!
$endgroup$
– Daman deep
Feb 3 at 11:51
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Write as
$$
U=frac{-ln X}{-ln X-ln (1-Y)}
$$
As you noted $-ln Xsim exp(1)$ and because $1-Ysim text{Unif}(0,1)$ it folows that $-ln(1-Y)sim exp(1)$. Since $-ln X$ and $-ln (1-Y)$ are independent it follows that $Usim text{Beta}(1,1)$ which is the same as a uniform random variable on $(0,1)$.
answered Feb 2 at 18:49
Foobaz JohnFoobaz John
22.9k41552
$endgroup$
$begingroup$
I was stuck on this for so long. Thanks a lot mate.
$endgroup$
– Vishesh Chugh
Feb 3 at 2:09
2
$begingroup$
@VisheshChugh Sorry but if you were stuck before, you should still be stuck since this answer does not show you how to prove the result but merely invokes a more general fact (which you probably do not know how to prove either) to deduce this one. Strange world...
$endgroup$
– Did
Feb 3 at 8:57
$begingroup$
@Did There's a thing known as confusion and that shouldn't be so hard for you to understand as you so much already. Don't know why you'd be so irrelevant. Strange world. :) :)
$endgroup$
– Vishesh Chugh
Feb 14 at 9:57
add a comment |
$begingroup$
Write as
$$
U=frac{-ln X}{-ln X-ln (1-Y)}
$$
As you noted $-ln Xsim exp(1)$ and because $1-Ysim text{Unif}(0,1)$ it folows that $-ln(1-Y)sim exp(1)$. Since $-ln X$ and $-ln (1-Y)$ are independent it follows that $Usim text{Beta}(1,1)$ which is the same as a uniform random variable on $(0,1)$.
answered Feb 2 at 18:49
Foobaz JohnFoobaz John
22.9k41552
$endgroup$
$begingroup$
I was stuck on this for so long. Thanks a lot mate.
$endgroup$
– Vishesh Chugh
Feb 3 at 2:09
2
$begingroup$
@VisheshChugh Sorry but if you were stuck before, you should still be stuck since this answer does not show you how to prove the result but merely invokes a more general fact (which you probably do not know how to prove either) to deduce this one. Strange world...
$endgroup$
– Did
Feb 3 at 8:57
$begingroup$
@Did There's a thing known as confusion and that shouldn't be so hard for you to understand as you so much already. Don't know why you'd be so irrelevant. Strange world. :) :)
$endgroup$
– Vishesh Chugh
Feb 14 at 9:57
add a comment |
$begingroup$
Write as
$$
U=frac{-ln X}{-ln X-ln (1-Y)}
$$
As you noted $-ln Xsim exp(1)$ and because $1-Ysim text{Unif}(0,1)$ it folows that $-ln(1-Y)sim exp(1)$. Since $-ln X$ and $-ln (1-Y)$ are independent it follows that $Usim text{Beta}(1,1)$ which is the same as a uniform random variable on $(0,1)$.
answered Feb 2 at 18:49
Foobaz JohnFoobaz John
22.9k41552
$endgroup$
Write as
$$
U=frac{-ln X}{-ln X-ln (1-Y)}
$$
As you noted $-ln Xsim exp(1)$ and because $1-Ysim text{Unif}(0,1)$ it folows that $-ln(1-Y)sim exp(1)$. Since $-ln X$ and $-ln (1-Y)$ are independent it follows that $Usim text{Beta}(1,1)$ which is the same as a uniform random variable on $(0,1)$.
answered Feb 2 at 18:49
Foobaz JohnFoobaz John
22.9k41552
answered Feb 2 at 18:49
Foobaz JohnFoobaz John
22.9k41552
answered Feb 2 at 18:49
Foobaz JohnFoobaz John
22.9k41552
answered Feb 2 at 18:49
Foobaz JohnFoobaz John
22.9k41552
22.9k41552
$begingroup$
I was stuck on this for so long. Thanks a lot mate.
$endgroup$
– Vishesh Chugh
Feb 3 at 2:09
2
$begingroup$
@VisheshChugh Sorry but if you were stuck before, you should still be stuck since this answer does not show you how to prove the result but merely invokes a more general fact (which you probably do not know how to prove either) to deduce this one. Strange world...
$endgroup$
– Did
Feb 3 at 8:57
$begingroup$
@Did There's a thing known as confusion and that shouldn't be so hard for you to understand as you so much already. Don't know why you'd be so irrelevant. Strange world. :) :)
$endgroup$
– Vishesh Chugh
Feb 14 at 9:57
add a comment |
$begingroup$
I was stuck on this for so long. Thanks a lot mate.
$endgroup$
– Vishesh Chugh
Feb 3 at 2:09
2
$begingroup$
@VisheshChugh Sorry but if you were stuck before, you should still be stuck since this answer does not show you how to prove the result but merely invokes a more general fact (which you probably do not know how to prove either) to deduce this one. Strange world...
$endgroup$
– Did
Feb 3 at 8:57
$begingroup$
@Did There's a thing known as confusion and that shouldn't be so hard for you to understand as you so much already. Don't know why you'd be so irrelevant. Strange world. :) :)
$endgroup$
– Vishesh Chugh
Feb 14 at 9:57
$begingroup$
I was stuck on this for so long. Thanks a lot mate.
$endgroup$
– Vishesh Chugh
Feb 3 at 2:09
$begingroup$
I was stuck on this for so long. Thanks a lot mate.
$endgroup$
– Vishesh Chugh
Feb 3 at 2:09
2
2
$begingroup$
@VisheshChugh Sorry but if you were stuck before, you should still be stuck since this answer does not show you how to prove the result but merely invokes a more general fact (which you probably do not know how to prove either) to deduce this one. Strange world...
$endgroup$
– Did
Feb 3 at 8:57
$begingroup$
@VisheshChugh Sorry but if you were stuck before, you should still be stuck since this answer does not show you how to prove the result but merely invokes a more general fact (which you probably do not know how to prove either) to deduce this one. Strange world...
$endgroup$
– Did
Feb 3 at 8:57
$begingroup$
@Did There's a thing known as confusion and that shouldn't be so hard for you to understand as you so much already. Don't know why you'd be so irrelevant. Strange world. :) :)
$endgroup$
– Vishesh Chugh
Feb 14 at 9:57
$begingroup$
@Did There's a thing known as confusion and that shouldn't be so hard for you to understand as you so much already. Don't know why you'd be so irrelevant. Strange world. :) :)
$endgroup$
– Vishesh Chugh
Feb 14 at 9:57
add a comment |
$begingroup$
$G(lambda)=e^{-x}x^{lambda-1};x>0,lambda>0 $ and $B_1(u,v)={x^{u-1}}(1-x)^{v-1} ; 0<x<1 ,u>0,v>0 $
$X sim G(lambda) $ and $Ysim G(m)$
$dfrac{X}{X+Y} sim B_1(lambda,m)$
And good luck for JAM $2019$
answered Feb 3 at 3:41
Daman deepDaman deep
756420
$endgroup$
$begingroup$
Thanks mate. Haha how do you know? You taking it too? This year?
$endgroup$
– Vishesh Chugh
Feb 3 at 10:51
$begingroup$
This question came in any previous year, I solved it few days back!
$endgroup$
– Daman deep
Feb 3 at 11:51
add a comment |
$begingroup$
$G(lambda)=e^{-x}x^{lambda-1};x>0,lambda>0 $ and $B_1(u,v)={x^{u-1}}(1-x)^{v-1} ; 0<x<1 ,u>0,v>0 $
$X sim G(lambda) $ and $Ysim G(m)$
$dfrac{X}{X+Y} sim B_1(lambda,m)$
And good luck for JAM $2019$
answered Feb 3 at 3:41
Daman deepDaman deep
756420
$endgroup$
$begingroup$
Thanks mate. Haha how do you know? You taking it too? This year?
$endgroup$
– Vishesh Chugh
Feb 3 at 10:51
$begingroup$
This question came in any previous year, I solved it few days back!
$endgroup$
– Daman deep
Feb 3 at 11:51
add a comment |
$begingroup$
$G(lambda)=e^{-x}x^{lambda-1};x>0,lambda>0 $ and $B_1(u,v)={x^{u-1}}(1-x)^{v-1} ; 0<x<1 ,u>0,v>0 $
$X sim G(lambda) $ and $Ysim G(m)$
$dfrac{X}{X+Y} sim B_1(lambda,m)$
And good luck for JAM $2019$
answered Feb 3 at 3:41
Daman deepDaman deep
756420
$endgroup$
$G(lambda)=e^{-x}x^{lambda-1};x>0,lambda>0 $ and $B_1(u,v)={x^{u-1}}(1-x)^{v-1} ; 0<x<1 ,u>0,v>0 $
$X sim G(lambda) $ and $Ysim G(m)$
$dfrac{X}{X+Y} sim B_1(lambda,m)$
And good luck for JAM $2019$
answered Feb 3 at 3:41
Daman deepDaman deep
756420
answered Feb 3 at 3:41
Daman deepDaman deep
756420
answered Feb 3 at 3:41
Daman deepDaman deep
756420
answered Feb 3 at 3:41
Daman deepDaman deep
756420
756420
$begingroup$
Thanks mate. Haha how do you know? You taking it too? This year?
$endgroup$
– Vishesh Chugh
Feb 3 at 10:51
$begingroup$
This question came in any previous year, I solved it few days back!
$endgroup$
– Daman deep
Feb 3 at 11:51
add a comment |
$begingroup$
Thanks mate. Haha how do you know? You taking it too? This year?
$endgroup$
– Vishesh Chugh
Feb 3 at 10:51
$begingroup$
This question came in any previous year, I solved it few days back!
$endgroup$
– Daman deep
Feb 3 at 11:51
$begingroup$
Thanks mate. Haha how do you know? You taking it too? This year?
$endgroup$
– Vishesh Chugh
Feb 3 at 10:51
$begingroup$
Thanks mate. Haha how do you know? You taking it too? This year?
$endgroup$
– Vishesh Chugh
Feb 3 at 10:51
$begingroup$
This question came in any previous year, I solved it few days back!
$endgroup$
– Daman deep
Feb 3 at 11:51
$begingroup$
This question came in any previous year, I solved it few days back!
$endgroup$
– Daman deep
Feb 3 at 11:51
add a comment |
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1
$begingroup$
You wrote "and $ln(1-Y)$ is also $U(0,1)$. But this is not true. Write your ratio as $S/(S+T)$ where $S,T$ are iid exponential rvs.
$endgroup$
– kimchi lover
Feb 2 at 18:08
1
$begingroup$
@kimchilover - presumably "and also $-ln(1-Y) sim exp(1)$" or something similar was intended
$endgroup$
– Henry
Feb 2 at 18:13
$begingroup$
@Henry I agree.
$endgroup$
– kimchi lover
Feb 2 at 18:15
1
$begingroup$
You will see $U$ has a standard distribution if you derive it (say, by a change of variables or using well-known relations between distributions), from which the variance follows immediately.
$endgroup$
– StubbornAtom
Feb 2 at 18:34
1
$begingroup$
@Vishesh You are not showing your work/attempts, so it's hard to know where you are stuck.
$endgroup$
– StubbornAtom
Feb 2 at 18:39