Mixing
The pdf of sum of -log($U_i$) in which Ui is iid uniform distributed
$begingroup$
Suppose $Ui$ is independently uniformed distributed between [0,b], $Y = -Sigma_1^n log(U_i)$. what is the pdf of Y? I tried used characteristic function but it doesn't match each of usual distribution.
probability-theory uniform-distribution
asked Jan 19 at 8:54
T.yT.y
82
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add a comment |
$begingroup$
Suppose $Ui$ is independently uniformed distributed between [0,b], $Y = -Sigma_1^n log(U_i)$. what is the pdf of Y? I tried used characteristic function but it doesn't match each of usual distribution.
probability-theory uniform-distribution
asked Jan 19 at 8:54
T.yT.y
82
$endgroup$
add a comment |
$begingroup$
Suppose $Ui$ is independently uniformed distributed between [0,b], $Y = -Sigma_1^n log(U_i)$. what is the pdf of Y? I tried used characteristic function but it doesn't match each of usual distribution.
probability-theory uniform-distribution
asked Jan 19 at 8:54
T.yT.y
82
$endgroup$
Suppose $Ui$ is independently uniformed distributed between [0,b], $Y = -Sigma_1^n log(U_i)$. what is the pdf of Y? I tried used characteristic function but it doesn't match each of usual distribution.
probability-theory uniform-distribution
probability-theory uniform-distribution
asked Jan 19 at 8:54
T.yT.y
82
asked Jan 19 at 8:54
T.yT.y
82
asked Jan 19 at 8:54
T.yT.y
82
asked Jan 19 at 8:54
T.yT.y
82
asked Jan 19 at 8:54
T.yT.y
82
82
add a comment |
add a comment |
1 Answer
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$begingroup$
Note: the following argument assumes $b=1$. To generalise, add $ln b$ to each $ln U_i$ term, i.e. $-nln b$ to $y$ so its pdf shifts.
You probably already worked out $-ln U_isimoperatorname{Exp}(1)$, because $$P(-ln Ule x)=P(Ugeexp -x)=1-exp -x.$$Of course, this implies $-ln U_i$ has characteristic function $1/(1-it)$, so $Y$ has cf $1/(1-it)^n$. Now, what distribution is that? Spoiler: it's
a Gamma distribution with $k=n,,theta=1$, so the pdf is $frac{y^{n-1}}{(n-1)!}exp -y$ for $yge 0$.
If you can guess that by thinking about integrals, the inversion theorem guarantees it's right because it gives the desired cf.
answered Jan 19 at 9:03
J.G.J.G.
28.1k22844
$endgroup$
$begingroup$
Thanks. I have already worked out the case when $b=1$, however I couldn't figure out the generalised part. Why can we just add $-nlnb$ to y? How can you prove it is right regularization?
$endgroup$
– T.y
Jan 19 at 9:11
$begingroup$
@T.y Because $U_isim U(0,,b)$ iff $U_i/bsim U(0,,1)$.
$endgroup$
– J.G.
Jan 19 at 9:13
$begingroup$
Still confused. I mean I already worked out the cf of general b version and it is %(b^(nit))/((1-it)^n)%. How can I match the pdf?
$endgroup$
– T.y
Jan 19 at 9:30
$begingroup$
@T.y Scaling $U_i$ to $bU_i$ changes the characteristic function $varphi(t)$ to $varphi(bt)$, not $b^{it}varphi(t)$.
$endgroup$
– J.G.
Jan 19 at 9:32
add a comment |
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$begingroup$
Note: the following argument assumes $b=1$. To generalise, add $ln b$ to each $ln U_i$ term, i.e. $-nln b$ to $y$ so its pdf shifts.
You probably already worked out $-ln U_isimoperatorname{Exp}(1)$, because $$P(-ln Ule x)=P(Ugeexp -x)=1-exp -x.$$Of course, this implies $-ln U_i$ has characteristic function $1/(1-it)$, so $Y$ has cf $1/(1-it)^n$. Now, what distribution is that? Spoiler: it's
a Gamma distribution with $k=n,,theta=1$, so the pdf is $frac{y^{n-1}}{(n-1)!}exp -y$ for $yge 0$.
If you can guess that by thinking about integrals, the inversion theorem guarantees it's right because it gives the desired cf.
answered Jan 19 at 9:03
J.G.J.G.
28.1k22844
$endgroup$
$begingroup$
Thanks. I have already worked out the case when $b=1$, however I couldn't figure out the generalised part. Why can we just add $-nlnb$ to y? How can you prove it is right regularization?
$endgroup$
– T.y
Jan 19 at 9:11
$begingroup$
@T.y Because $U_isim U(0,,b)$ iff $U_i/bsim U(0,,1)$.
$endgroup$
– J.G.
Jan 19 at 9:13
$begingroup$
Still confused. I mean I already worked out the cf of general b version and it is %(b^(nit))/((1-it)^n)%. How can I match the pdf?
$endgroup$
– T.y
Jan 19 at 9:30
$begingroup$
@T.y Scaling $U_i$ to $bU_i$ changes the characteristic function $varphi(t)$ to $varphi(bt)$, not $b^{it}varphi(t)$.
$endgroup$
– J.G.
Jan 19 at 9:32
add a comment |
$begingroup$
Note: the following argument assumes $b=1$. To generalise, add $ln b$ to each $ln U_i$ term, i.e. $-nln b$ to $y$ so its pdf shifts.
You probably already worked out $-ln U_isimoperatorname{Exp}(1)$, because $$P(-ln Ule x)=P(Ugeexp -x)=1-exp -x.$$Of course, this implies $-ln U_i$ has characteristic function $1/(1-it)$, so $Y$ has cf $1/(1-it)^n$. Now, what distribution is that? Spoiler: it's
a Gamma distribution with $k=n,,theta=1$, so the pdf is $frac{y^{n-1}}{(n-1)!}exp -y$ for $yge 0$.
If you can guess that by thinking about integrals, the inversion theorem guarantees it's right because it gives the desired cf.
answered Jan 19 at 9:03
J.G.J.G.
28.1k22844
$endgroup$
$begingroup$
Thanks. I have already worked out the case when $b=1$, however I couldn't figure out the generalised part. Why can we just add $-nlnb$ to y? How can you prove it is right regularization?
$endgroup$
– T.y
Jan 19 at 9:11
$begingroup$
@T.y Because $U_isim U(0,,b)$ iff $U_i/bsim U(0,,1)$.
$endgroup$
– J.G.
Jan 19 at 9:13
$begingroup$
Still confused. I mean I already worked out the cf of general b version and it is %(b^(nit))/((1-it)^n)%. How can I match the pdf?
$endgroup$
– T.y
Jan 19 at 9:30
$begingroup$
@T.y Scaling $U_i$ to $bU_i$ changes the characteristic function $varphi(t)$ to $varphi(bt)$, not $b^{it}varphi(t)$.
$endgroup$
– J.G.
Jan 19 at 9:32
add a comment |
$begingroup$
Note: the following argument assumes $b=1$. To generalise, add $ln b$ to each $ln U_i$ term, i.e. $-nln b$ to $y$ so its pdf shifts.
You probably already worked out $-ln U_isimoperatorname{Exp}(1)$, because $$P(-ln Ule x)=P(Ugeexp -x)=1-exp -x.$$Of course, this implies $-ln U_i$ has characteristic function $1/(1-it)$, so $Y$ has cf $1/(1-it)^n$. Now, what distribution is that? Spoiler: it's
a Gamma distribution with $k=n,,theta=1$, so the pdf is $frac{y^{n-1}}{(n-1)!}exp -y$ for $yge 0$.
If you can guess that by thinking about integrals, the inversion theorem guarantees it's right because it gives the desired cf.
answered Jan 19 at 9:03
J.G.J.G.
28.1k22844
$endgroup$
Note: the following argument assumes $b=1$. To generalise, add $ln b$ to each $ln U_i$ term, i.e. $-nln b$ to $y$ so its pdf shifts.
You probably already worked out $-ln U_isimoperatorname{Exp}(1)$, because $$P(-ln Ule x)=P(Ugeexp -x)=1-exp -x.$$Of course, this implies $-ln U_i$ has characteristic function $1/(1-it)$, so $Y$ has cf $1/(1-it)^n$. Now, what distribution is that? Spoiler: it's
a Gamma distribution with $k=n,,theta=1$, so the pdf is $frac{y^{n-1}}{(n-1)!}exp -y$ for $yge 0$.
If you can guess that by thinking about integrals, the inversion theorem guarantees it's right because it gives the desired cf.
answered Jan 19 at 9:03
J.G.J.G.
28.1k22844
answered Jan 19 at 9:03
J.G.J.G.
28.1k22844
answered Jan 19 at 9:03
J.G.J.G.
28.1k22844
answered Jan 19 at 9:03
J.G.J.G.
28.1k22844
28.1k22844
$begingroup$
Thanks. I have already worked out the case when $b=1$, however I couldn't figure out the generalised part. Why can we just add $-nlnb$ to y? How can you prove it is right regularization?
$endgroup$
– T.y
Jan 19 at 9:11
$begingroup$
@T.y Because $U_isim U(0,,b)$ iff $U_i/bsim U(0,,1)$.
$endgroup$
– J.G.
Jan 19 at 9:13
$begingroup$
Still confused. I mean I already worked out the cf of general b version and it is %(b^(nit))/((1-it)^n)%. How can I match the pdf?
$endgroup$
– T.y
Jan 19 at 9:30
$begingroup$
@T.y Scaling $U_i$ to $bU_i$ changes the characteristic function $varphi(t)$ to $varphi(bt)$, not $b^{it}varphi(t)$.
$endgroup$
– J.G.
Jan 19 at 9:32
add a comment |
$begingroup$
Thanks. I have already worked out the case when $b=1$, however I couldn't figure out the generalised part. Why can we just add $-nlnb$ to y? How can you prove it is right regularization?
$endgroup$
– T.y
Jan 19 at 9:11
$begingroup$
@T.y Because $U_isim U(0,,b)$ iff $U_i/bsim U(0,,1)$.
$endgroup$
– J.G.
Jan 19 at 9:13
$begingroup$
Still confused. I mean I already worked out the cf of general b version and it is %(b^(nit))/((1-it)^n)%. How can I match the pdf?
$endgroup$
– T.y
Jan 19 at 9:30
$begingroup$
@T.y Scaling $U_i$ to $bU_i$ changes the characteristic function $varphi(t)$ to $varphi(bt)$, not $b^{it}varphi(t)$.
$endgroup$
– J.G.
Jan 19 at 9:32
$begingroup$
Thanks. I have already worked out the case when $b=1$, however I couldn't figure out the generalised part. Why can we just add $-nlnb$ to y? How can you prove it is right regularization?
$endgroup$
– T.y
Jan 19 at 9:11
$begingroup$
Thanks. I have already worked out the case when $b=1$, however I couldn't figure out the generalised part. Why can we just add $-nlnb$ to y? How can you prove it is right regularization?
$endgroup$
– T.y
Jan 19 at 9:11
$begingroup$
@T.y Because $U_isim U(0,,b)$ iff $U_i/bsim U(0,,1)$.
$endgroup$
– J.G.
Jan 19 at 9:13
$begingroup$
@T.y Because $U_isim U(0,,b)$ iff $U_i/bsim U(0,,1)$.
$endgroup$
– J.G.
Jan 19 at 9:13
$begingroup$
Still confused. I mean I already worked out the cf of general b version and it is %(b^(nit))/((1-it)^n)%. How can I match the pdf?
$endgroup$
– T.y
Jan 19 at 9:30
$begingroup$
Still confused. I mean I already worked out the cf of general b version and it is %(b^(nit))/((1-it)^n)%. How can I match the pdf?
$endgroup$
– T.y
Jan 19 at 9:30
$begingroup$
@T.y Scaling $U_i$ to $bU_i$ changes the characteristic function $varphi(t)$ to $varphi(bt)$, not $b^{it}varphi(t)$.
$endgroup$
– J.G.
Jan 19 at 9:32
$begingroup$
@T.y Scaling $U_i$ to $bU_i$ changes the characteristic function $varphi(t)$ to $varphi(bt)$, not $b^{it}varphi(t)$.
$endgroup$
– J.G.
Jan 19 at 9:32
add a comment |
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