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Decomposing a discrete random variable

Constructive proof for a problem of discrete random variables.Representing random variables in a shared and private decompositionProof of convergence in distribution of a discrete random variableShowing that ${rm E}[X]=sum_{k=0}^infty P(X>k)$ for a discrete random variableRandom Variable depending on another Random Variable?Question about probability of a random variable with uniform distributionEasiest way to recognize a discrete random variable distributionExpected Value of Random Variable with ParameterProbability density of transformed random variableWhat is a sample of a random variable?Making Random variable from other Random variables but keep them independentConstructive proof for a problem of discrete random variables.

0

1

$begingroup$

Note: this question is a duplicate of another question asked by my classmate with much more contexts. Attempt to suggest edits is rejected due to drastic change, so I reask the question.

---

I hate to admit it, but I'm asking for help on a homework problem assigned in our Information Theory class. I'm really scratching my head but unable to prove it:

For two discrete random variables $X, Y sim p(x,y)$, we can find another random variable $Z$ independent of $X$, such that there exists a function $f$ satisfying $Y = f(X,Z)$.

The problem also comes with a hint that calls for a constructive proof.

Now that I have (maybe in a wrong way) constructed a random variable $Y'$ that has the same distribution as $Y$, but I simply can't prove that $Y = Y'$ due to their identical distribution.

This question I found may be related to the problem but it goes unanswered. Any help is appreciated!

probability-theory random-variables information-theory

share|cite|improve this question

asked Mar 20 at 13:34

skyfiretimesskyfiretimes

12

$endgroup$

add a comment | 

0

1

$begingroup$

Note: this question is a duplicate of another question asked by my classmate with much more contexts. Attempt to suggest edits is rejected due to drastic change, so I reask the question.

---

I hate to admit it, but I'm asking for help on a homework problem assigned in our Information Theory class. I'm really scratching my head but unable to prove it:

For two discrete random variables $X, Y sim p(x,y)$, we can find another random variable $Z$ independent of $X$, such that there exists a function $f$ satisfying $Y = f(X,Z)$.

The problem also comes with a hint that calls for a constructive proof.

Now that I have (maybe in a wrong way) constructed a random variable $Y'$ that has the same distribution as $Y$, but I simply can't prove that $Y = Y'$ due to their identical distribution.

This question I found may be related to the problem but it goes unanswered. Any help is appreciated!

probability-theory random-variables information-theory

share|cite|improve this question

asked Mar 20 at 13:34

skyfiretimesskyfiretimes

12

$endgroup$

add a comment | 

$begingroup$

Note: this question is a duplicate of another question asked by my classmate with much more contexts. Attempt to suggest edits is rejected due to drastic change, so I reask the question.

---

I hate to admit it, but I'm asking for help on a homework problem assigned in our Information Theory class. I'm really scratching my head but unable to prove it:

For two discrete random variables $X, Y sim p(x,y)$, we can find another random variable $Z$ independent of $X$, such that there exists a function $f$ satisfying $Y = f(X,Z)$.

The problem also comes with a hint that calls for a constructive proof.

Now that I have (maybe in a wrong way) constructed a random variable $Y'$ that has the same distribution as $Y$, but I simply can't prove that $Y = Y'$ due to their identical distribution.

This question I found may be related to the problem but it goes unanswered. Any help is appreciated!

probability-theory random-variables information-theory

share|cite|improve this question

asked Mar 20 at 13:34

skyfiretimesskyfiretimes

12

$endgroup$

share|cite|improve this question

asked Mar 20 at 13:34

skyfiretimesskyfiretimes

12

share|cite|improve this question

asked Mar 20 at 13:34

skyfiretimesskyfiretimes

12

asked Mar 20 at 13:34

skyfiretimesskyfiretimes

12

asked Mar 20 at 13:34

skyfiretimesskyfiretimes

12

3 Answers
3

active

oldest

votes

0

$begingroup$

Given
$X = x$ and $Y = y$, let $Z$ be uniform on the interval $left[mathbb P(Y < y mid X=x), mathbb P(Y le y mid X = x)right]$. Then $Y = min {y: mathbb P(Y le y mid X) ge Z}$. That is, when $X = x$ and $Z = z in (0,1)$, $Y = min {y: mathbb P(Y le y mid X = x) ge z }$.

share|cite|improve this answer

answered Mar 20 at 14:09

Robert IsraelRobert Israel

330k23220475

$endgroup$

add a comment | 

0

$begingroup$

Let $Z$ be uniform on $[0,1]$. Given $X=x$, use inverse transform sampling to generate $Y$, using $Z$ as the uniform variable and using $q(y)= p(x,y)/left(sum_y p(x,y)right)$ as the distribution.

share|cite|improve this answer

answered Mar 20 at 18:06

Mike EarnestMike Earnest

27.1k22152

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If I undestand inverse transform sampling correctly, I can recover a random variable by the inverse of its CDF, so is this just an alternative way of constructing a random variable with same distribution as $Y$?
  $endgroup$
  – skyfiretimes
  Mar 21 at 14:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes. If $F$ is a cdf, and $Zsim text{Unif}(0,1)$, then the random variable $W$ defined by $W=min{winmathbb R:F(w)ge Z}$ will have $F$ as its cdf. Here, we are using this to construct a random variable whose distribution is the conditional distribution of $Y$ given ${X=x}$. So, letting $F_x(y)=P(Yle y|X=x)$, the formula for $Y$ is $$Y=min{yinmathbb R:F_x(y)ge Z}.$$
  $endgroup$
  – Mike Earnest
  Mar 21 at 14:49
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh I see the problem; I have constructed something which is equal in distribution to $Y$, but not equal to $Y$. Let me think more.@skyfiretimes
  $endgroup$
  – Mike Earnest
  Mar 21 at 14:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yep, that's the difficult part about this problem. Thanks for your answer anyway :)
  $endgroup$
  – skyfiretimes
  Mar 21 at 14:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I do not think $f(X,Z)=Y$ is possible, only $f(X,Z)stackrel{d}=Y$. Consider this small example; there are three cards with the labels $(0,0),(0,1)$ and $(1,0)$. Choose a random card; $X$ is the first coordinate, $Y$ is the second. In order to have $Y=f(X,Z)$, you need $Z$ to be a function of which card was picked. But that probability space is so small, that there does not even exist a random variable $Z$ which is independent of $X$ (except $Z$ constant). @skyfiretimes
  $endgroup$
  – Mike Earnest
  Mar 21 at 15:30
  

  
  
  
  

add a comment | 

0

$begingroup$

I'll post my professor's answer here. The problem is called Functional Representataion Lemma and you may search for its proof on the internet. The conclusion of the lemma is actually stronger, though.

share|cite|improve this answer

answered Mar 23 at 2:47

skyfiretimesskyfiretimes

12

$endgroup$

add a comment | 

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0

$begingroup$

Given
$X = x$ and $Y = y$, let $Z$ be uniform on the interval $left[mathbb P(Y < y mid X=x), mathbb P(Y le y mid X = x)right]$. Then $Y = min {y: mathbb P(Y le y mid X) ge Z}$. That is, when $X = x$ and $Z = z in (0,1)$, $Y = min {y: mathbb P(Y le y mid X = x) ge z }$.

share|cite|improve this answer

answered Mar 20 at 14:09

Robert IsraelRobert Israel

330k23220475

$endgroup$

add a comment | 

0

$begingroup$

Given
$X = x$ and $Y = y$, let $Z$ be uniform on the interval $left[mathbb P(Y < y mid X=x), mathbb P(Y le y mid X = x)right]$. Then $Y = min {y: mathbb P(Y le y mid X) ge Z}$. That is, when $X = x$ and $Z = z in (0,1)$, $Y = min {y: mathbb P(Y le y mid X = x) ge z }$.

share|cite|improve this answer

answered Mar 20 at 14:09

Robert IsraelRobert Israel

330k23220475

$endgroup$

add a comment | 

$begingroup$

Given
$X = x$ and $Y = y$, let $Z$ be uniform on the interval $left[mathbb P(Y < y mid X=x), mathbb P(Y le y mid X = x)right]$. Then $Y = min {y: mathbb P(Y le y mid X) ge Z}$. That is, when $X = x$ and $Z = z in (0,1)$, $Y = min {y: mathbb P(Y le y mid X = x) ge z }$.

share|cite|improve this answer

answered Mar 20 at 14:09

Robert IsraelRobert Israel

330k23220475

$endgroup$

share|cite|improve this answer

answered Mar 20 at 14:09

Robert IsraelRobert Israel

330k23220475

answered Mar 20 at 14:09

Robert IsraelRobert Israel

330k23220475

answered Mar 20 at 14:09

Robert IsraelRobert Israel

330k23220475

0

$begingroup$

Let $Z$ be uniform on $[0,1]$. Given $X=x$, use inverse transform sampling to generate $Y$, using $Z$ as the uniform variable and using $q(y)= p(x,y)/left(sum_y p(x,y)right)$ as the distribution.

share|cite|improve this answer

answered Mar 20 at 18:06

Mike EarnestMike Earnest

27.1k22152

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If I undestand inverse transform sampling correctly, I can recover a random variable by the inverse of its CDF, so is this just an alternative way of constructing a random variable with same distribution as $Y$?
  $endgroup$
  – skyfiretimes
  Mar 21 at 14:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes. If $F$ is a cdf, and $Zsim text{Unif}(0,1)$, then the random variable $W$ defined by $W=min{winmathbb R:F(w)ge Z}$ will have $F$ as its cdf. Here, we are using this to construct a random variable whose distribution is the conditional distribution of $Y$ given ${X=x}$. So, letting $F_x(y)=P(Yle y|X=x)$, the formula for $Y$ is $$Y=min{yinmathbb R:F_x(y)ge Z}.$$
  $endgroup$
  – Mike Earnest
  Mar 21 at 14:49
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh I see the problem; I have constructed something which is equal in distribution to $Y$, but not equal to $Y$. Let me think more.@skyfiretimes
  $endgroup$
  – Mike Earnest
  Mar 21 at 14:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yep, that's the difficult part about this problem. Thanks for your answer anyway :)
  $endgroup$
  – skyfiretimes
  Mar 21 at 14:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I do not think $f(X,Z)=Y$ is possible, only $f(X,Z)stackrel{d}=Y$. Consider this small example; there are three cards with the labels $(0,0),(0,1)$ and $(1,0)$. Choose a random card; $X$ is the first coordinate, $Y$ is the second. In order to have $Y=f(X,Z)$, you need $Z$ to be a function of which card was picked. But that probability space is so small, that there does not even exist a random variable $Z$ which is independent of $X$ (except $Z$ constant). @skyfiretimes
  $endgroup$
  – Mike Earnest
  Mar 21 at 15:30
  

  
  
  
  

add a comment | 

0

$begingroup$

Let $Z$ be uniform on $[0,1]$. Given $X=x$, use inverse transform sampling to generate $Y$, using $Z$ as the uniform variable and using $q(y)= p(x,y)/left(sum_y p(x,y)right)$ as the distribution.

share|cite|improve this answer

answered Mar 20 at 18:06

Mike EarnestMike Earnest

27.1k22152

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If I undestand inverse transform sampling correctly, I can recover a random variable by the inverse of its CDF, so is this just an alternative way of constructing a random variable with same distribution as $Y$?
  $endgroup$
  – skyfiretimes
  Mar 21 at 14:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes. If $F$ is a cdf, and $Zsim text{Unif}(0,1)$, then the random variable $W$ defined by $W=min{winmathbb R:F(w)ge Z}$ will have $F$ as its cdf. Here, we are using this to construct a random variable whose distribution is the conditional distribution of $Y$ given ${X=x}$. So, letting $F_x(y)=P(Yle y|X=x)$, the formula for $Y$ is $$Y=min{yinmathbb R:F_x(y)ge Z}.$$
  $endgroup$
  – Mike Earnest
  Mar 21 at 14:49
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh I see the problem; I have constructed something which is equal in distribution to $Y$, but not equal to $Y$. Let me think more.@skyfiretimes
  $endgroup$
  – Mike Earnest
  Mar 21 at 14:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yep, that's the difficult part about this problem. Thanks for your answer anyway :)
  $endgroup$
  – skyfiretimes
  Mar 21 at 14:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I do not think $f(X,Z)=Y$ is possible, only $f(X,Z)stackrel{d}=Y$. Consider this small example; there are three cards with the labels $(0,0),(0,1)$ and $(1,0)$. Choose a random card; $X$ is the first coordinate, $Y$ is the second. In order to have $Y=f(X,Z)$, you need $Z$ to be a function of which card was picked. But that probability space is so small, that there does not even exist a random variable $Z$ which is independent of $X$ (except $Z$ constant). @skyfiretimes
  $endgroup$
  – Mike Earnest
  Mar 21 at 15:30
  

  
  
  
  

add a comment | 

$begingroup$

Let $Z$ be uniform on $[0,1]$. Given $X=x$, use inverse transform sampling to generate $Y$, using $Z$ as the uniform variable and using $q(y)= p(x,y)/left(sum_y p(x,y)right)$ as the distribution.

share|cite|improve this answer

answered Mar 20 at 18:06

Mike EarnestMike Earnest

27.1k22152

$endgroup$

share|cite|improve this answer

answered Mar 20 at 18:06

Mike EarnestMike Earnest

27.1k22152

answered Mar 20 at 18:06

Mike EarnestMike Earnest

27.1k22152

answered Mar 20 at 18:06

Mike EarnestMike Earnest

27.1k22152

0

$begingroup$

I'll post my professor's answer here. The problem is called Functional Representataion Lemma and you may search for its proof on the internet. The conclusion of the lemma is actually stronger, though.

share|cite|improve this answer

answered Mar 23 at 2:47

skyfiretimesskyfiretimes

12

$endgroup$

add a comment | 

0

$begingroup$

I'll post my professor's answer here. The problem is called Functional Representataion Lemma and you may search for its proof on the internet. The conclusion of the lemma is actually stronger, though.

share|cite|improve this answer

answered Mar 23 at 2:47

skyfiretimesskyfiretimes

12

$endgroup$

add a comment | 

$begingroup$

I'll post my professor's answer here. The problem is called Functional Representataion Lemma and you may search for its proof on the internet. The conclusion of the lemma is actually stronger, though.

share|cite|improve this answer

answered Mar 23 at 2:47

skyfiretimesskyfiretimes

12

$endgroup$

share|cite|improve this answer

answered Mar 23 at 2:47

skyfiretimesskyfiretimes

12

answered Mar 23 at 2:47

skyfiretimesskyfiretimes

12

answered Mar 23 at 2:47

skyfiretimesskyfiretimes

12