Stochastic Independence of $tan(U_1)$ and $tan(U_1+U_2)$ for uniform independent $U_1,U_2$
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I found the following statement in Stoyanov's book on counterexamples (Section 7.2) quite interesting: Let $U_1$ and $U_2$ be independent and uniformly distributed on $(0,pi)$. Then $tan(U_1)$ and $tan(U_1+U_2)$ are independent.
This is taken from https://projecteuclid.org/download/pdf_1/euclid.aoms/1177698885 and I am really confused by that: To my mind, they say in the last two sections that $U_1+U_2$ is also uniformly distributed and independent of $U_1$ (which would imply the statement) - this is clearly not true so that I must have misunderstood.
I would appreciate a clarification of this and I would like to know in what sense and why there actually is independence between $tan(U_1)$ and $tan(U_1+U_2)$.
probability convolution independence uniform-distribution
asked yesterday
Mau314
36418
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up vote
3
down vote
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I found the following statement in Stoyanov's book on counterexamples (Section 7.2) quite interesting: Let $U_1$ and $U_2$ be independent and uniformly distributed on $(0,pi)$. Then $tan(U_1)$ and $tan(U_1+U_2)$ are independent.
This is taken from https://projecteuclid.org/download/pdf_1/euclid.aoms/1177698885 and I am really confused by that: To my mind, they say in the last two sections that $U_1+U_2$ is also uniformly distributed and independent of $U_1$ (which would imply the statement) - this is clearly not true so that I must have misunderstood.
I would appreciate a clarification of this and I would like to know in what sense and why there actually is independence between $tan(U_1)$ and $tan(U_1+U_2)$.
probability convolution independence uniform-distribution
asked yesterday
Mau314
36418
This is interesting! $tan $ is not one-to -one on $(0,2pi)$ so you cannot say that independence of $tan (U_1)$ and $tan (U_1+U_2)$ implies that of $U_1$ and $U_1+U_2$. Note that $tan (U_1)$ is not even defined when $U_1=pi /2$ but this event has probability $0$. The weird behavior an $tan $ may lead to this result and I don't have a proof right now.
– Kavi Rama Murthy
yesterday
@KaviRamaMurthy Do you think this might be extended to other functions which are periodic in $(0, 2pi)$ and odd with respect to $pi$? Very interesting indeed.
– Lee David Chung Lin
yesterday
2
The statements there should be saying that $P$ has an arbitrary distribution, and $Q sim text{Uniform}(0, pi)$, independent of $P$. Then $[(P + Q) mod pi] sim text{Uniform}(0, pi)$ and it is independent from $P$. The mod part convention is stated in the last sentence in P. 2 (P. 917).
– BGM
yesterday
@BGM That's much better, thanks. The proof is okay if $P$ is also uniform, I'll also try to understand the general case.
– Mau314
23 hours ago
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I found the following statement in Stoyanov's book on counterexamples (Section 7.2) quite interesting: Let $U_1$ and $U_2$ be independent and uniformly distributed on $(0,pi)$. Then $tan(U_1)$ and $tan(U_1+U_2)$ are independent.
This is taken from https://projecteuclid.org/download/pdf_1/euclid.aoms/1177698885 and I am really confused by that: To my mind, they say in the last two sections that $U_1+U_2$ is also uniformly distributed and independent of $U_1$ (which would imply the statement) - this is clearly not true so that I must have misunderstood.
I would appreciate a clarification of this and I would like to know in what sense and why there actually is independence between $tan(U_1)$ and $tan(U_1+U_2)$.
probability convolution independence uniform-distribution
asked yesterday
Mau314
36418
I found the following statement in Stoyanov's book on counterexamples (Section 7.2) quite interesting: Let $U_1$ and $U_2$ be independent and uniformly distributed on $(0,pi)$. Then $tan(U_1)$ and $tan(U_1+U_2)$ are independent.
This is taken from https://projecteuclid.org/download/pdf_1/euclid.aoms/1177698885 and I am really confused by that: To my mind, they say in the last two sections that $U_1+U_2$ is also uniformly distributed and independent of $U_1$ (which would imply the statement) - this is clearly not true so that I must have misunderstood.
I would appreciate a clarification of this and I would like to know in what sense and why there actually is independence between $tan(U_1)$ and $tan(U_1+U_2)$.
probability convolution independence uniform-distribution
probability convolution independence uniform-distribution
asked yesterday
Mau314
36418
asked yesterday
Mau314
36418
asked yesterday
Mau314
36418
asked yesterday
Mau314
36418
asked yesterday
Mau314
36418
36418
This is interesting! $tan $ is not one-to -one on $(0,2pi)$ so you cannot say that independence of $tan (U_1)$ and $tan (U_1+U_2)$ implies that of $U_1$ and $U_1+U_2$. Note that $tan (U_1)$ is not even defined when $U_1=pi /2$ but this event has probability $0$. The weird behavior an $tan $ may lead to this result and I don't have a proof right now.
– Kavi Rama Murthy
yesterday
@KaviRamaMurthy Do you think this might be extended to other functions which are periodic in $(0, 2pi)$ and odd with respect to $pi$? Very interesting indeed.
– Lee David Chung Lin
yesterday
2
The statements there should be saying that $P$ has an arbitrary distribution, and $Q sim text{Uniform}(0, pi)$, independent of $P$. Then $[(P + Q) mod pi] sim text{Uniform}(0, pi)$ and it is independent from $P$. The mod part convention is stated in the last sentence in P. 2 (P. 917).
– BGM
yesterday
@BGM That's much better, thanks. The proof is okay if $P$ is also uniform, I'll also try to understand the general case.
– Mau314
23 hours ago
add a comment |
This is interesting! $tan $ is not one-to -one on $(0,2pi)$ so you cannot say that independence of $tan (U_1)$ and $tan (U_1+U_2)$ implies that of $U_1$ and $U_1+U_2$. Note that $tan (U_1)$ is not even defined when $U_1=pi /2$ but this event has probability $0$. The weird behavior an $tan $ may lead to this result and I don't have a proof right now.
– Kavi Rama Murthy
yesterday
@KaviRamaMurthy Do you think this might be extended to other functions which are periodic in $(0, 2pi)$ and odd with respect to $pi$? Very interesting indeed.
– Lee David Chung Lin
yesterday
2
The statements there should be saying that $P$ has an arbitrary distribution, and $Q sim text{Uniform}(0, pi)$, independent of $P$. Then $[(P + Q) mod pi] sim text{Uniform}(0, pi)$ and it is independent from $P$. The mod part convention is stated in the last sentence in P. 2 (P. 917).
– BGM
yesterday
@BGM That's much better, thanks. The proof is okay if $P$ is also uniform, I'll also try to understand the general case.
– Mau314
23 hours ago
This is interesting! $tan $ is not one-to -one on $(0,2pi)$ so you cannot say that independence of $tan (U_1)$ and $tan (U_1+U_2)$ implies that of $U_1$ and $U_1+U_2$. Note that $tan (U_1)$ is not even defined when $U_1=pi /2$ but this event has probability $0$. The weird behavior an $tan $ may lead to this result and I don't have a proof right now.
– Kavi Rama Murthy
yesterday
This is interesting! $tan $ is not one-to -one on $(0,2pi)$ so you cannot say that independence of $tan (U_1)$ and $tan (U_1+U_2)$ implies that of $U_1$ and $U_1+U_2$. Note that $tan (U_1)$ is not even defined when $U_1=pi /2$ but this event has probability $0$. The weird behavior an $tan $ may lead to this result and I don't have a proof right now.
– Kavi Rama Murthy
yesterday
@KaviRamaMurthy Do you think this might be extended to other functions which are periodic in $(0, 2pi)$ and odd with respect to $pi$? Very interesting indeed.
– Lee David Chung Lin
yesterday
@KaviRamaMurthy Do you think this might be extended to other functions which are periodic in $(0, 2pi)$ and odd with respect to $pi$? Very interesting indeed.
– Lee David Chung Lin
yesterday
2
2
The statements there should be saying that $P$ has an arbitrary distribution, and $Q sim text{Uniform}(0, pi)$, independent of $P$. Then $[(P + Q) mod pi] sim text{Uniform}(0, pi)$ and it is independent from $P$. The mod part convention is stated in the last sentence in P. 2 (P. 917).
– BGM
yesterday
The statements there should be saying that $P$ has an arbitrary distribution, and $Q sim text{Uniform}(0, pi)$, independent of $P$. Then $[(P + Q) mod pi] sim text{Uniform}(0, pi)$ and it is independent from $P$. The mod part convention is stated in the last sentence in P. 2 (P. 917).
– BGM
yesterday
@BGM That's much better, thanks. The proof is okay if $P$ is also uniform, I'll also try to understand the general case.
– Mau314
23 hours ago
@BGM That's much better, thanks. The proof is okay if $P$ is also uniform, I'll also try to understand the general case.
– Mau314
23 hours ago
add a comment |
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This is interesting! $tan $ is not one-to -one on $(0,2pi)$ so you cannot say that independence of $tan (U_1)$ and $tan (U_1+U_2)$ implies that of $U_1$ and $U_1+U_2$. Note that $tan (U_1)$ is not even defined when $U_1=pi /2$ but this event has probability $0$. The weird behavior an $tan $ may lead to this result and I don't have a proof right now.
– Kavi Rama Murthy
yesterday
@KaviRamaMurthy Do you think this might be extended to other functions which are periodic in $(0, 2pi)$ and odd with respect to $pi$? Very interesting indeed.
– Lee David Chung Lin
yesterday
2
The statements there should be saying that $P$ has an arbitrary distribution, and $Q sim text{Uniform}(0, pi)$, independent of $P$. Then $[(P + Q) mod pi] sim text{Uniform}(0, pi)$ and it is independent from $P$. The mod part convention is stated in the last sentence in P. 2 (P. 917).
– BGM
yesterday
@BGM That's much better, thanks. The proof is okay if $P$ is also uniform, I'll also try to understand the general case.
– Mau314
23 hours ago