Conditional expectation of Binomial variable The Next CEO of Stack OverflowConditional...

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Conditional expectation of Binomial variable



The Next CEO of Stack OverflowConditional expectation equals random variable almost sureCalculate the conditional expectation for a uniformly distributed variableConditional expectation of product of independent RVsGeometric distribution and expectation of polynomials of random variableProbability of Mixture of Normal Random VariablesDistribution of Conditional Bernoulli Random VariableCDF of a mixed stochastic variableCalculation with conditional expectation.A question about conditional expectation problem from DurrettSymmetricity of binomial distribution












0












$begingroup$


Let $ X sim U(70,100) $ be a discrete distribution such that $ P{X=i} = frac{1}{31} $ and $ Y sim B(i,0.75) $



What is the $ E[Y] $ ?



I don't understand the solution which is



$ E[Y] = E[E[Y|X]] = E[0.75X] = 0.75E[X] = 0.75 cdot frac{100+70}{2} = 63.75$



I don't understand how to get to this answer. I try:



$$ E[Y] = E[E[Y|X]] = sum_x E[Y| X=x] cdot P{X=x} = sum_x y cdot P{Y=y, X=x} =\ sum_x y cdot P{Y=y}cdot P{X=x} = y cdot P{Y=y} cdot sum_x P{X=x} = y cdot P{Y=y} $$






probability-theory







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Your first line makes no sense. Did you mean to say "A random pair $(X, Y)$ is distributed in such a way that $X sim mathrm{Unif}(70, 100)$ and given $X = i,$ $Y sim mathrm{Bin}(i, 3/5)$"?
    $endgroup$
    – Will M.
    Mar 17 at 18:15








  • 1




    $begingroup$
    Anyway, remember that the expectation of binomial $(n, p)$ is $np,$ thus, the expectation of $Y$ given $X$ is $3/4X.$
    $endgroup$
    – Will M.
    Mar 17 at 18:18










  • $begingroup$
    Yes, correct that is what I meant. Thanks. Could you also explain to me why $ E[XY|X] = XE[Y|X] $ ? I'm trying to find $ E[XY] $. I'm talking about those two $ X, Y $ mentioned above.
    $endgroup$
    – bm1125
    Mar 17 at 18:26












  • $begingroup$
    I am not sure if you know measure theory or not, but intuitively, given $X,$ we know handle it as a deterministic quantity, hence $XY$ given $X$ is like $cY$ and so $E(cY) = cE(Y).$
    $endgroup$
    – Will M.
    Mar 17 at 18:44










  • $begingroup$
    Oh that’s because $ X is like a parameter ?
    $endgroup$
    – bm1125
    Mar 17 at 18:56
















0












$begingroup$


Let $ X sim U(70,100) $ be a discrete distribution such that $ P{X=i} = frac{1}{31} $ and $ Y sim B(i,0.75) $



What is the $ E[Y] $ ?



I don't understand the solution which is



$ E[Y] = E[E[Y|X]] = E[0.75X] = 0.75E[X] = 0.75 cdot frac{100+70}{2} = 63.75$



I don't understand how to get to this answer. I try:



$$ E[Y] = E[E[Y|X]] = sum_x E[Y| X=x] cdot P{X=x} = sum_x y cdot P{Y=y, X=x} =\ sum_x y cdot P{Y=y}cdot P{X=x} = y cdot P{Y=y} cdot sum_x P{X=x} = y cdot P{Y=y} $$






probability-theory







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Your first line makes no sense. Did you mean to say "A random pair $(X, Y)$ is distributed in such a way that $X sim mathrm{Unif}(70, 100)$ and given $X = i,$ $Y sim mathrm{Bin}(i, 3/5)$"?
    $endgroup$
    – Will M.
    Mar 17 at 18:15








  • 1




    $begingroup$
    Anyway, remember that the expectation of binomial $(n, p)$ is $np,$ thus, the expectation of $Y$ given $X$ is $3/4X.$
    $endgroup$
    – Will M.
    Mar 17 at 18:18










  • $begingroup$
    Yes, correct that is what I meant. Thanks. Could you also explain to me why $ E[XY|X] = XE[Y|X] $ ? I'm trying to find $ E[XY] $. I'm talking about those two $ X, Y $ mentioned above.
    $endgroup$
    – bm1125
    Mar 17 at 18:26












  • $begingroup$
    I am not sure if you know measure theory or not, but intuitively, given $X,$ we know handle it as a deterministic quantity, hence $XY$ given $X$ is like $cY$ and so $E(cY) = cE(Y).$
    $endgroup$
    – Will M.
    Mar 17 at 18:44










  • $begingroup$
    Oh that’s because $ X is like a parameter ?
    $endgroup$
    – bm1125
    Mar 17 at 18:56














0












0








0





$begingroup$


Let $ X sim U(70,100) $ be a discrete distribution such that $ P{X=i} = frac{1}{31} $ and $ Y sim B(i,0.75) $



What is the $ E[Y] $ ?



I don't understand the solution which is



$ E[Y] = E[E[Y|X]] = E[0.75X] = 0.75E[X] = 0.75 cdot frac{100+70}{2} = 63.75$



I don't understand how to get to this answer. I try:



$$ E[Y] = E[E[Y|X]] = sum_x E[Y| X=x] cdot P{X=x} = sum_x y cdot P{Y=y, X=x} =\ sum_x y cdot P{Y=y}cdot P{X=x} = y cdot P{Y=y} cdot sum_x P{X=x} = y cdot P{Y=y} $$






probability-theory







share|cite|improve this question











$endgroup$




Let $ X sim U(70,100) $ be a discrete distribution such that $ P{X=i} = frac{1}{31} $ and $ Y sim B(i,0.75) $



What is the $ E[Y] $ ?



I don't understand the solution which is



$ E[Y] = E[E[Y|X]] = E[0.75X] = 0.75E[X] = 0.75 cdot frac{100+70}{2} = 63.75$



I don't understand how to get to this answer. I try:



$$ E[Y] = E[E[Y|X]] = sum_x E[Y| X=x] cdot P{X=x} = sum_x y cdot P{Y=y, X=x} =\ sum_x y cdot P{Y=y}cdot P{X=x} = y cdot P{Y=y} cdot sum_x P{X=x} = y cdot P{Y=y} $$






probability-theory




probability-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 17 at 17:53







bm1125

















asked Mar 17 at 17:21









bm1125bm1125

67816




67816








  • 1




    $begingroup$
    Your first line makes no sense. Did you mean to say "A random pair $(X, Y)$ is distributed in such a way that $X sim mathrm{Unif}(70, 100)$ and given $X = i,$ $Y sim mathrm{Bin}(i, 3/5)$"?
    $endgroup$
    – Will M.
    Mar 17 at 18:15








  • 1




    $begingroup$
    Anyway, remember that the expectation of binomial $(n, p)$ is $np,$ thus, the expectation of $Y$ given $X$ is $3/4X.$
    $endgroup$
    – Will M.
    Mar 17 at 18:18










  • $begingroup$
    Yes, correct that is what I meant. Thanks. Could you also explain to me why $ E[XY|X] = XE[Y|X] $ ? I'm trying to find $ E[XY] $. I'm talking about those two $ X, Y $ mentioned above.
    $endgroup$
    – bm1125
    Mar 17 at 18:26












  • $begingroup$
    I am not sure if you know measure theory or not, but intuitively, given $X,$ we know handle it as a deterministic quantity, hence $XY$ given $X$ is like $cY$ and so $E(cY) = cE(Y).$
    $endgroup$
    – Will M.
    Mar 17 at 18:44










  • $begingroup$
    Oh that’s because $ X is like a parameter ?
    $endgroup$
    – bm1125
    Mar 17 at 18:56














  • 1




    $begingroup$
    Your first line makes no sense. Did you mean to say "A random pair $(X, Y)$ is distributed in such a way that $X sim mathrm{Unif}(70, 100)$ and given $X = i,$ $Y sim mathrm{Bin}(i, 3/5)$"?
    $endgroup$
    – Will M.
    Mar 17 at 18:15








  • 1




    $begingroup$
    Anyway, remember that the expectation of binomial $(n, p)$ is $np,$ thus, the expectation of $Y$ given $X$ is $3/4X.$
    $endgroup$
    – Will M.
    Mar 17 at 18:18










  • $begingroup$
    Yes, correct that is what I meant. Thanks. Could you also explain to me why $ E[XY|X] = XE[Y|X] $ ? I'm trying to find $ E[XY] $. I'm talking about those two $ X, Y $ mentioned above.
    $endgroup$
    – bm1125
    Mar 17 at 18:26












  • $begingroup$
    I am not sure if you know measure theory or not, but intuitively, given $X,$ we know handle it as a deterministic quantity, hence $XY$ given $X$ is like $cY$ and so $E(cY) = cE(Y).$
    $endgroup$
    – Will M.
    Mar 17 at 18:44










  • $begingroup$
    Oh that’s because $ X is like a parameter ?
    $endgroup$
    – bm1125
    Mar 17 at 18:56








1




1




$begingroup$
Your first line makes no sense. Did you mean to say "A random pair $(X, Y)$ is distributed in such a way that $X sim mathrm{Unif}(70, 100)$ and given $X = i,$ $Y sim mathrm{Bin}(i, 3/5)$"?
$endgroup$
– Will M.
Mar 17 at 18:15






$begingroup$
Your first line makes no sense. Did you mean to say "A random pair $(X, Y)$ is distributed in such a way that $X sim mathrm{Unif}(70, 100)$ and given $X = i,$ $Y sim mathrm{Bin}(i, 3/5)$"?
$endgroup$
– Will M.
Mar 17 at 18:15






1




1




$begingroup$
Anyway, remember that the expectation of binomial $(n, p)$ is $np,$ thus, the expectation of $Y$ given $X$ is $3/4X.$
$endgroup$
– Will M.
Mar 17 at 18:18




$begingroup$
Anyway, remember that the expectation of binomial $(n, p)$ is $np,$ thus, the expectation of $Y$ given $X$ is $3/4X.$
$endgroup$
– Will M.
Mar 17 at 18:18












$begingroup$
Yes, correct that is what I meant. Thanks. Could you also explain to me why $ E[XY|X] = XE[Y|X] $ ? I'm trying to find $ E[XY] $. I'm talking about those two $ X, Y $ mentioned above.
$endgroup$
– bm1125
Mar 17 at 18:26






$begingroup$
Yes, correct that is what I meant. Thanks. Could you also explain to me why $ E[XY|X] = XE[Y|X] $ ? I'm trying to find $ E[XY] $. I'm talking about those two $ X, Y $ mentioned above.
$endgroup$
– bm1125
Mar 17 at 18:26














$begingroup$
I am not sure if you know measure theory or not, but intuitively, given $X,$ we know handle it as a deterministic quantity, hence $XY$ given $X$ is like $cY$ and so $E(cY) = cE(Y).$
$endgroup$
– Will M.
Mar 17 at 18:44




$begingroup$
I am not sure if you know measure theory or not, but intuitively, given $X,$ we know handle it as a deterministic quantity, hence $XY$ given $X$ is like $cY$ and so $E(cY) = cE(Y).$
$endgroup$
– Will M.
Mar 17 at 18:44












$begingroup$
Oh that’s because $ X is like a parameter ?
$endgroup$
– bm1125
Mar 17 at 18:56




$begingroup$
Oh that’s because $ X is like a parameter ?
$endgroup$
– bm1125
Mar 17 at 18:56










1 Answer
1






active

oldest

votes


















1












$begingroup$

Your answer started true but continued wrong! Hence your answer, we can write:$$E[Y] {= E[E[Y|X]] \= sum_x E[Y| X=x] cdot P{X=x}\=sum_x E[Y| X=x] cdot {1over 31}\=sum_{x=70}^{100} 0.75cdot x cdot {1over 31}\={3over 124}sum_{x=70}^{100}x\.\.\.\=63.75}$$






share|cite|improve this answer









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    oldest

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    $begingroup$

    Your answer started true but continued wrong! Hence your answer, we can write:$$E[Y] {= E[E[Y|X]] \= sum_x E[Y| X=x] cdot P{X=x}\=sum_x E[Y| X=x] cdot {1over 31}\=sum_{x=70}^{100} 0.75cdot x cdot {1over 31}\={3over 124}sum_{x=70}^{100}x\.\.\.\=63.75}$$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Your answer started true but continued wrong! Hence your answer, we can write:$$E[Y] {= E[E[Y|X]] \= sum_x E[Y| X=x] cdot P{X=x}\=sum_x E[Y| X=x] cdot {1over 31}\=sum_{x=70}^{100} 0.75cdot x cdot {1over 31}\={3over 124}sum_{x=70}^{100}x\.\.\.\=63.75}$$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Your answer started true but continued wrong! Hence your answer, we can write:$$E[Y] {= E[E[Y|X]] \= sum_x E[Y| X=x] cdot P{X=x}\=sum_x E[Y| X=x] cdot {1over 31}\=sum_{x=70}^{100} 0.75cdot x cdot {1over 31}\={3over 124}sum_{x=70}^{100}x\.\.\.\=63.75}$$






        share|cite|improve this answer









        $endgroup$



        Your answer started true but continued wrong! Hence your answer, we can write:$$E[Y] {= E[E[Y|X]] \= sum_x E[Y| X=x] cdot P{X=x}\=sum_x E[Y| X=x] cdot {1over 31}\=sum_{x=70}^{100} 0.75cdot x cdot {1over 31}\={3over 124}sum_{x=70}^{100}x\.\.\.\=63.75}$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Mar 17 at 20:55









        Mostafa AyazMostafa Ayaz

        18.3k31040




        18.3k31040






























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