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Hard expected value problem

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0

$begingroup$

Given a series of discrete random variables $Y_2, Y_3...Y_n$, such that for all $Y_i$ :

$P(Y_i = e^i) = frac1i$, $P(Y_i = 3) = frac13$, $P(Y_i = X) = frac23 - frac1i$,

and $X$ is a random variable such that $P(X = 1) = frac23$, $P(x = -1) = frac13$. find $mathbbE(Y_i)$ in terms of $i$.

I tried quite a lot of approaches to solve this but no success yet. First, I acknowledged that $P(Y_i = -1) + P(Y_i = 1) = P(Y_i =X)$, but I can't find another equation and so I can't find all of the probabilities for some $Y_i$.

probability random-variables expected-value

share|cite|improve this question

edited Mar 22 at 0:01

Jacob Jones

14111

asked Mar 21 at 23:50

stos1512stos1512

716

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Can you clarify the different values that $Y_i$ can take?
  $endgroup$
  – Akash Patel
  Mar 21 at 23:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  e^i, 3, 1 and -1
  $endgroup$
  – stos1512
  Mar 22 at 0:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Side observation: Especially with one of the possible values being $e^i$, $i$ is an unfortunate choice of a dummy variable...
  $endgroup$
  – Brian Tung
  Mar 22 at 0:15
  

  
  
  
  

add a comment | 

0

$begingroup$

Given a series of discrete random variables $Y_2, Y_3...Y_n$, such that for all $Y_i$ :

$P(Y_i = e^i) = frac1i$, $P(Y_i = 3) = frac13$, $P(Y_i = X) = frac23 - frac1i$,

and $X$ is a random variable such that $P(X = 1) = frac23$, $P(x = -1) = frac13$. find $mathbbE(Y_i)$ in terms of $i$.

I tried quite a lot of approaches to solve this but no success yet. First, I acknowledged that $P(Y_i = -1) + P(Y_i = 1) = P(Y_i =X)$, but I can't find another equation and so I can't find all of the probabilities for some $Y_i$.

probability random-variables expected-value

share|cite|improve this question

edited Mar 22 at 0:01

Jacob Jones

14111

asked Mar 21 at 23:50

stos1512stos1512

716

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Can you clarify the different values that $Y_i$ can take?
  $endgroup$
  – Akash Patel
  Mar 21 at 23:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  e^i, 3, 1 and -1
  $endgroup$
  – stos1512
  Mar 22 at 0:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Side observation: Especially with one of the possible values being $e^i$, $i$ is an unfortunate choice of a dummy variable...
  $endgroup$
  – Brian Tung
  Mar 22 at 0:15
  

  
  
  
  

add a comment | 

$begingroup$

Given a series of discrete random variables $Y_2, Y_3...Y_n$, such that for all $Y_i$ :

$P(Y_i = e^i) = frac1i$, $P(Y_i = 3) = frac13$, $P(Y_i = X) = frac23 - frac1i$,

and $X$ is a random variable such that $P(X = 1) = frac23$, $P(x = -1) = frac13$. find $mathbbE(Y_i)$ in terms of $i$.

I tried quite a lot of approaches to solve this but no success yet. First, I acknowledged that $P(Y_i = -1) + P(Y_i = 1) = P(Y_i =X)$, but I can't find another equation and so I can't find all of the probabilities for some $Y_i$.

probability random-variables expected-value

share|cite|improve this question

edited Mar 22 at 0:01

Jacob Jones

14111

asked Mar 21 at 23:50

stos1512stos1512

716

$endgroup$

share|cite|improve this question

edited Mar 22 at 0:01

Jacob Jones

14111

asked Mar 21 at 23:50

stos1512stos1512

716

share|cite|improve this question

edited Mar 22 at 0:01

Jacob Jones

14111

asked Mar 21 at 23:50

stos1512stos1512

716

edited Mar 22 at 0:01

Jacob Jones

14111

edited Mar 22 at 0:01

Jacob Jones

14111

asked Mar 21 at 23:50

stos1512stos1512

716

asked Mar 21 at 23:50

stos1512stos1512

716

2 Answers
2

active

oldest

votes

0

$begingroup$

Use the Law of Total Expectation: $beginalignBbb E(Y_i) = Bbb E(Bbb E(Y_imid X)) endalign$

For clarrity, let us define $mathsf P_i(y):=Pr(Y_i=y)$

$$beginalignBbb E(Y_i) &= Bbb E(Bbb E(Y_imid X))\[1ex]&= Bbb Ebig(e^imathsf P_i(e^i)+3mathsf P_i(3)+Xmathsf P_i(X)big)\[1ex]&= e^imathsf P_i(e^i)+3mathsf P_i(3)+Bbb Ebig(Xmathsf P_i(X)big)\[1ex]&~~vdots endalign$$

share|cite|improve this answer

edited Mar 22 at 0:14

answered Mar 22 at 0:03

Graham KempGraham Kemp

87.8k43578

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  im sorry' but i cant understand how you got the inner conditional expectation. I reckon the probabilities should be the conditional ones, and if so- how did you decide they are just equal to the original probabilities ?
  $endgroup$
  – stos1512
  Mar 22 at 0:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For a given value of $X$, we see that $Y$ may realise the values $e^i, 3, X$ with the provided probabilities, $mathsf P_i(e^i),mathsf P_i(3),mathsf P_i(X)$, so therefore the conditional expectation for $Y$ given $X$ is...
  $endgroup$
  – Graham Kemp
  Mar 22 at 0:19
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Alternatively you can see for $xin-1,1$ we have $Pr(Y=x)=Pr(Y=Xmid X=x)cdotPr(X=x)$.
  $endgroup$
  – Graham Kemp
  Mar 22 at 0:34
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I don't know what im missing here, i still dont see why the probabilities from the inner conditional expectation, that Y = e^i and that Y = 3 shouldnt be conditional probabilities, given that X = x. How did you come to realize that they are equal ?
  $endgroup$
  – stos1512
  Mar 22 at 9:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  could you maybe elaborate on why theses conditional probabilities are equal to the original ones ?
  $endgroup$
  – stos1512
  Mar 22 at 9:41
  

  
  
  
  

 | 
show 2 more comments

0

$begingroup$

We remove $X$ from this by saying that $P(Y_i=X)= frac23-frac1i$ is the same as saying $P(Y_i=1)= frac23(frac23-frac1i)$ and $P(Y_i=-1)= frac13(frac23-frac1i)$

Then, $E[Y_i]=y_iP(Y_i=y_i)=e^i(frac1i)+3(frac13)+1(frac23(frac23-frac1i))-1(frac13(frac23-frac1i))=frace^ii+1+frac49-frac23i-frac29+frac13i=frace^ii+frac119-frac13i$

share|cite|improve this answer

answered Mar 22 at 0:20

Akash PatelAkash Patel

168110

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  why ? this is true if Yi = X and Yi = 1 are independent isnt it ?..which doesnt seem like it from a first glance.
  $endgroup$
  – stos1512
  Mar 22 at 9:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This follows from $P(Y=x)=P(Y=X|X=x)P(X=x)$
  $endgroup$
  – Akash Patel
  Mar 22 at 15:28
  

  
  
  
  

add a comment | 

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0

$begingroup$

Use the Law of Total Expectation: $beginalignBbb E(Y_i) = Bbb E(Bbb E(Y_imid X)) endalign$

For clarrity, let us define $mathsf P_i(y):=Pr(Y_i=y)$

$$beginalignBbb E(Y_i) &= Bbb E(Bbb E(Y_imid X))\[1ex]&= Bbb Ebig(e^imathsf P_i(e^i)+3mathsf P_i(3)+Xmathsf P_i(X)big)\[1ex]&= e^imathsf P_i(e^i)+3mathsf P_i(3)+Bbb Ebig(Xmathsf P_i(X)big)\[1ex]&~~vdots endalign$$

share|cite|improve this answer

edited Mar 22 at 0:14

answered Mar 22 at 0:03

Graham KempGraham Kemp

87.8k43578

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  im sorry' but i cant understand how you got the inner conditional expectation. I reckon the probabilities should be the conditional ones, and if so- how did you decide they are just equal to the original probabilities ?
  $endgroup$
  – stos1512
  Mar 22 at 0:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For a given value of $X$, we see that $Y$ may realise the values $e^i, 3, X$ with the provided probabilities, $mathsf P_i(e^i),mathsf P_i(3),mathsf P_i(X)$, so therefore the conditional expectation for $Y$ given $X$ is...
  $endgroup$
  – Graham Kemp
  Mar 22 at 0:19
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Alternatively you can see for $xin-1,1$ we have $Pr(Y=x)=Pr(Y=Xmid X=x)cdotPr(X=x)$.
  $endgroup$
  – Graham Kemp
  Mar 22 at 0:34
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I don't know what im missing here, i still dont see why the probabilities from the inner conditional expectation, that Y = e^i and that Y = 3 shouldnt be conditional probabilities, given that X = x. How did you come to realize that they are equal ?
  $endgroup$
  – stos1512
  Mar 22 at 9:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  could you maybe elaborate on why theses conditional probabilities are equal to the original ones ?
  $endgroup$
  – stos1512
  Mar 22 at 9:41
  

  
  
  
  

 | 
show 2 more comments

0

$begingroup$

Use the Law of Total Expectation: $beginalignBbb E(Y_i) = Bbb E(Bbb E(Y_imid X)) endalign$

For clarrity, let us define $mathsf P_i(y):=Pr(Y_i=y)$

$$beginalignBbb E(Y_i) &= Bbb E(Bbb E(Y_imid X))\[1ex]&= Bbb Ebig(e^imathsf P_i(e^i)+3mathsf P_i(3)+Xmathsf P_i(X)big)\[1ex]&= e^imathsf P_i(e^i)+3mathsf P_i(3)+Bbb Ebig(Xmathsf P_i(X)big)\[1ex]&~~vdots endalign$$

share|cite|improve this answer

edited Mar 22 at 0:14

answered Mar 22 at 0:03

Graham KempGraham Kemp

87.8k43578

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  im sorry' but i cant understand how you got the inner conditional expectation. I reckon the probabilities should be the conditional ones, and if so- how did you decide they are just equal to the original probabilities ?
  $endgroup$
  – stos1512
  Mar 22 at 0:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For a given value of $X$, we see that $Y$ may realise the values $e^i, 3, X$ with the provided probabilities, $mathsf P_i(e^i),mathsf P_i(3),mathsf P_i(X)$, so therefore the conditional expectation for $Y$ given $X$ is...
  $endgroup$
  – Graham Kemp
  Mar 22 at 0:19
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Alternatively you can see for $xin-1,1$ we have $Pr(Y=x)=Pr(Y=Xmid X=x)cdotPr(X=x)$.
  $endgroup$
  – Graham Kemp
  Mar 22 at 0:34
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I don't know what im missing here, i still dont see why the probabilities from the inner conditional expectation, that Y = e^i and that Y = 3 shouldnt be conditional probabilities, given that X = x. How did you come to realize that they are equal ?
  $endgroup$
  – stos1512
  Mar 22 at 9:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  could you maybe elaborate on why theses conditional probabilities are equal to the original ones ?
  $endgroup$
  – stos1512
  Mar 22 at 9:41
  

  
  
  
  

 | 
show 2 more comments

$begingroup$

Use the Law of Total Expectation: $beginalignBbb E(Y_i) = Bbb E(Bbb E(Y_imid X)) endalign$

For clarrity, let us define $mathsf P_i(y):=Pr(Y_i=y)$

$$beginalignBbb E(Y_i) &= Bbb E(Bbb E(Y_imid X))\[1ex]&= Bbb Ebig(e^imathsf P_i(e^i)+3mathsf P_i(3)+Xmathsf P_i(X)big)\[1ex]&= e^imathsf P_i(e^i)+3mathsf P_i(3)+Bbb Ebig(Xmathsf P_i(X)big)\[1ex]&~~vdots endalign$$

share|cite|improve this answer

edited Mar 22 at 0:14

answered Mar 22 at 0:03

Graham KempGraham Kemp

87.8k43578

$endgroup$

share|cite|improve this answer

edited Mar 22 at 0:14

answered Mar 22 at 0:03

Graham KempGraham Kemp

87.8k43578

answered Mar 22 at 0:03

Graham KempGraham Kemp

87.8k43578

answered Mar 22 at 0:03

Graham KempGraham Kemp

87.8k43578

0

$begingroup$

We remove $X$ from this by saying that $P(Y_i=X)= frac23-frac1i$ is the same as saying $P(Y_i=1)= frac23(frac23-frac1i)$ and $P(Y_i=-1)= frac13(frac23-frac1i)$

Then, $E[Y_i]=y_iP(Y_i=y_i)=e^i(frac1i)+3(frac13)+1(frac23(frac23-frac1i))-1(frac13(frac23-frac1i))=frace^ii+1+frac49-frac23i-frac29+frac13i=frace^ii+frac119-frac13i$

share|cite|improve this answer

answered Mar 22 at 0:20

Akash PatelAkash Patel

168110

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  why ? this is true if Yi = X and Yi = 1 are independent isnt it ?..which doesnt seem like it from a first glance.
  $endgroup$
  – stos1512
  Mar 22 at 9:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This follows from $P(Y=x)=P(Y=X|X=x)P(X=x)$
  $endgroup$
  – Akash Patel
  Mar 22 at 15:28
  

  
  
  
  

add a comment | 

0

$begingroup$

We remove $X$ from this by saying that $P(Y_i=X)= frac23-frac1i$ is the same as saying $P(Y_i=1)= frac23(frac23-frac1i)$ and $P(Y_i=-1)= frac13(frac23-frac1i)$

Then, $E[Y_i]=y_iP(Y_i=y_i)=e^i(frac1i)+3(frac13)+1(frac23(frac23-frac1i))-1(frac13(frac23-frac1i))=frace^ii+1+frac49-frac23i-frac29+frac13i=frace^ii+frac119-frac13i$

share|cite|improve this answer

answered Mar 22 at 0:20

Akash PatelAkash Patel

168110

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  why ? this is true if Yi = X and Yi = 1 are independent isnt it ?..which doesnt seem like it from a first glance.
  $endgroup$
  – stos1512
  Mar 22 at 9:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This follows from $P(Y=x)=P(Y=X|X=x)P(X=x)$
  $endgroup$
  – Akash Patel
  Mar 22 at 15:28
  

  
  
  
  

add a comment | 

$begingroup$

We remove $X$ from this by saying that $P(Y_i=X)= frac23-frac1i$ is the same as saying $P(Y_i=1)= frac23(frac23-frac1i)$ and $P(Y_i=-1)= frac13(frac23-frac1i)$

Then, $E[Y_i]=y_iP(Y_i=y_i)=e^i(frac1i)+3(frac13)+1(frac23(frac23-frac1i))-1(frac13(frac23-frac1i))=frace^ii+1+frac49-frac23i-frac29+frac13i=frace^ii+frac119-frac13i$

share|cite|improve this answer

answered Mar 22 at 0:20

Akash PatelAkash Patel

168110

$endgroup$

share|cite|improve this answer

answered Mar 22 at 0:20

Akash PatelAkash Patel

168110

answered Mar 22 at 0:20

Akash PatelAkash Patel

168110

answered Mar 22 at 0:20

Akash PatelAkash Patel

168110