X*Z has same distribution of Y*Z? [closed]
$begingroup$
Suppose I have three r.v.s $X, Y, Z$ such that $X$ and $Y$ are identically distributed.
Can we say that $Xcdot Z$ and $Ycdot Z$ have the same distribution? Can we prove it or disprove it?
Note: $X, Y, Z$ may be continuous r.v. and may be not necessarily be independent.
Thanks a lot in advance!
probability-theory
asked Dec 1 '18 at 10:43
GiulioGiulio
273
$endgroup$
closed as off-topic by amWhy, José Carlos Santos, user10354138, GNUSupporter 8964民主女神 地下教會, Paul Frost Dec 1 '18 at 11:31
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, José Carlos Santos, user10354138, GNUSupporter 8964民主女神 地下教會, Paul Frost
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Suppose I have three r.v.s $X, Y, Z$ such that $X$ and $Y$ are identically distributed.
Can we say that $Xcdot Z$ and $Ycdot Z$ have the same distribution? Can we prove it or disprove it?
Note: $X, Y, Z$ may be continuous r.v. and may be not necessarily be independent.
Thanks a lot in advance!
probability-theory
asked Dec 1 '18 at 10:43
GiulioGiulio
273
$endgroup$
closed as off-topic by amWhy, José Carlos Santos, user10354138, GNUSupporter 8964民主女神 地下教會, Paul Frost Dec 1 '18 at 11:31
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, José Carlos Santos, user10354138, GNUSupporter 8964民主女神 地下教會, Paul Frost
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Suppose I have three r.v.s $X, Y, Z$ such that $X$ and $Y$ are identically distributed.
Can we say that $Xcdot Z$ and $Ycdot Z$ have the same distribution? Can we prove it or disprove it?
Note: $X, Y, Z$ may be continuous r.v. and may be not necessarily be independent.
Thanks a lot in advance!
probability-theory
asked Dec 1 '18 at 10:43
GiulioGiulio
273
$endgroup$
Suppose I have three r.v.s $X, Y, Z$ such that $X$ and $Y$ are identically distributed.
Can we say that $Xcdot Z$ and $Ycdot Z$ have the same distribution? Can we prove it or disprove it?
Note: $X, Y, Z$ may be continuous r.v. and may be not necessarily be independent.
Thanks a lot in advance!
probability-theory
probability-theory
asked Dec 1 '18 at 10:43
GiulioGiulio
273
asked Dec 1 '18 at 10:43
GiulioGiulio
273
edited Dec 5 '18 at 20:54
Giulio
asked Dec 1 '18 at 10:43
GiulioGiulio
273
asked Dec 1 '18 at 10:43
GiulioGiulio
273
asked Dec 1 '18 at 10:43
GiulioGiulio
273
273
closed as off-topic by amWhy, José Carlos Santos, user10354138, GNUSupporter 8964民主女神 地下教會, Paul Frost Dec 1 '18 at 11:31
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, José Carlos Santos, user10354138, GNUSupporter 8964民主女神 地下教會, Paul Frost
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by amWhy, José Carlos Santos, user10354138, GNUSupporter 8964民主女神 地下教會, Paul Frost Dec 1 '18 at 11:31
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, José Carlos Santos, user10354138, GNUSupporter 8964民主女神 地下教會, Paul Frost
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Clearly they can have the same distribution, for example if all three are independent. But they do not have to have the same distribution
For example, let $X=1$ or $-1$ each with probability $frac12$, $Y=-X$ and $Z=X$. All three have the same distribution. Then $XZ=1$ with probability $1$ while $YZ=-1$ with probability $1$, so these are different
answered Dec 1 '18 at 10:52
HenryHenry
99.3k479165
$endgroup$
$begingroup$
Thanks a lot! It's a clever answer to a question I didn't realise it was so trivial.
$endgroup$
– Giulio
Dec 1 '18 at 10:57
$begingroup$
If they are independent, then they must have the same distribution. Is this what you mean?
$endgroup$
– MPW
Dec 1 '18 at 11:32
$begingroup$
@MPW - yes, though are also other examples where $XZ$ and $YZ$ have the same distribution, such as $X=Y$
$endgroup$
– Henry
Dec 1 '18 at 17:05
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Clearly they can have the same distribution, for example if all three are independent. But they do not have to have the same distribution
For example, let $X=1$ or $-1$ each with probability $frac12$, $Y=-X$ and $Z=X$. All three have the same distribution. Then $XZ=1$ with probability $1$ while $YZ=-1$ with probability $1$, so these are different
answered Dec 1 '18 at 10:52
HenryHenry
99.3k479165
$endgroup$
$begingroup$
Thanks a lot! It's a clever answer to a question I didn't realise it was so trivial.
$endgroup$
– Giulio
Dec 1 '18 at 10:57
$begingroup$
If they are independent, then they must have the same distribution. Is this what you mean?
$endgroup$
– MPW
Dec 1 '18 at 11:32
$begingroup$
@MPW - yes, though are also other examples where $XZ$ and $YZ$ have the same distribution, such as $X=Y$
$endgroup$
– Henry
Dec 1 '18 at 17:05
add a comment |
$begingroup$
Clearly they can have the same distribution, for example if all three are independent. But they do not have to have the same distribution
For example, let $X=1$ or $-1$ each with probability $frac12$, $Y=-X$ and $Z=X$. All three have the same distribution. Then $XZ=1$ with probability $1$ while $YZ=-1$ with probability $1$, so these are different
answered Dec 1 '18 at 10:52
HenryHenry
99.3k479165
$endgroup$
$begingroup$
Thanks a lot! It's a clever answer to a question I didn't realise it was so trivial.
$endgroup$
– Giulio
Dec 1 '18 at 10:57
$begingroup$
If they are independent, then they must have the same distribution. Is this what you mean?
$endgroup$
– MPW
Dec 1 '18 at 11:32
$begingroup$
@MPW - yes, though are also other examples where $XZ$ and $YZ$ have the same distribution, such as $X=Y$
$endgroup$
– Henry
Dec 1 '18 at 17:05
add a comment |
$begingroup$
Clearly they can have the same distribution, for example if all three are independent. But they do not have to have the same distribution
For example, let $X=1$ or $-1$ each with probability $frac12$, $Y=-X$ and $Z=X$. All three have the same distribution. Then $XZ=1$ with probability $1$ while $YZ=-1$ with probability $1$, so these are different
answered Dec 1 '18 at 10:52
HenryHenry
99.3k479165
$endgroup$
Clearly they can have the same distribution, for example if all three are independent. But they do not have to have the same distribution
For example, let $X=1$ or $-1$ each with probability $frac12$, $Y=-X$ and $Z=X$. All three have the same distribution. Then $XZ=1$ with probability $1$ while $YZ=-1$ with probability $1$, so these are different
answered Dec 1 '18 at 10:52
HenryHenry
99.3k479165
answered Dec 1 '18 at 10:52
HenryHenry
99.3k479165
answered Dec 1 '18 at 10:52
HenryHenry
99.3k479165
answered Dec 1 '18 at 10:52
HenryHenry
99.3k479165
99.3k479165
$begingroup$
Thanks a lot! It's a clever answer to a question I didn't realise it was so trivial.
$endgroup$
– Giulio
Dec 1 '18 at 10:57
$begingroup$
If they are independent, then they must have the same distribution. Is this what you mean?
$endgroup$
– MPW
Dec 1 '18 at 11:32
$begingroup$
@MPW - yes, though are also other examples where $XZ$ and $YZ$ have the same distribution, such as $X=Y$
$endgroup$
– Henry
Dec 1 '18 at 17:05
add a comment |
$begingroup$
Thanks a lot! It's a clever answer to a question I didn't realise it was so trivial.
$endgroup$
– Giulio
Dec 1 '18 at 10:57
$begingroup$
If they are independent, then they must have the same distribution. Is this what you mean?
$endgroup$
– MPW
Dec 1 '18 at 11:32
$begingroup$
@MPW - yes, though are also other examples where $XZ$ and $YZ$ have the same distribution, such as $X=Y$
$endgroup$
– Henry
Dec 1 '18 at 17:05
$begingroup$
Thanks a lot! It's a clever answer to a question I didn't realise it was so trivial.
$endgroup$
– Giulio
Dec 1 '18 at 10:57
$begingroup$
Thanks a lot! It's a clever answer to a question I didn't realise it was so trivial.
$endgroup$
– Giulio
Dec 1 '18 at 10:57
$begingroup$
If they are independent, then they must have the same distribution. Is this what you mean?
$endgroup$
– MPW
Dec 1 '18 at 11:32
$begingroup$
If they are independent, then they must have the same distribution. Is this what you mean?
$endgroup$
– MPW
Dec 1 '18 at 11:32
$begingroup$
@MPW - yes, though are also other examples where $XZ$ and $YZ$ have the same distribution, such as $X=Y$
$endgroup$
– Henry
Dec 1 '18 at 17:05
$begingroup$
@MPW - yes, though are also other examples where $XZ$ and $YZ$ have the same distribution, such as $X=Y$
$endgroup$
– Henry
Dec 1 '18 at 17:05
add a comment |
