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
$begingroup$
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
asked Mar 17 at 17:21
bm1125bm1125
67816
$endgroup$
add a comment |
$begingroup$
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
asked Mar 17 at 17:21
bm1125bm1125
67816
$endgroup$
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
add a comment |
$begingroup$
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
asked Mar 17 at 17:21
bm1125bm1125
67816
$endgroup$
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
asked Mar 17 at 17:21
bm1125bm1125
67816
asked Mar 17 at 17:21
bm1125bm1125
67816
edited Mar 17 at 17:53
bm1125
asked Mar 17 at 17:21
bm1125bm1125
67816
asked Mar 17 at 17:21
bm1125bm1125
67816
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your answer started true but continued wrong! Hence your answer, we can write:$$E[Y] {= E[E[Y|X]] \= sum_x E[Y| X=x] cdot P{X=x}\=sum_x E[Y| X=x] cdot {1over 31}\=sum_{x=70}^{100} 0.75cdot x cdot {1over 31}\={3over 124}sum_{x=70}^{100}x\.\.\.\=63.75}$$
answered Mar 17 at 20:55
Mostafa AyazMostafa Ayaz
18.3k31040
$endgroup$
add a comment |
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$begingroup$
Your answer started true but continued wrong! Hence your answer, we can write:$$E[Y] {= E[E[Y|X]] \= sum_x E[Y| X=x] cdot P{X=x}\=sum_x E[Y| X=x] cdot {1over 31}\=sum_{x=70}^{100} 0.75cdot x cdot {1over 31}\={3over 124}sum_{x=70}^{100}x\.\.\.\=63.75}$$
answered Mar 17 at 20:55
Mostafa AyazMostafa Ayaz
18.3k31040
$endgroup$
add a comment |
$begingroup$
Your answer started true but continued wrong! Hence your answer, we can write:$$E[Y] {= E[E[Y|X]] \= sum_x E[Y| X=x] cdot P{X=x}\=sum_x E[Y| X=x] cdot {1over 31}\=sum_{x=70}^{100} 0.75cdot x cdot {1over 31}\={3over 124}sum_{x=70}^{100}x\.\.\.\=63.75}$$
answered Mar 17 at 20:55
Mostafa AyazMostafa Ayaz
18.3k31040
$endgroup$
add a comment |
$begingroup$
Your answer started true but continued wrong! Hence your answer, we can write:$$E[Y] {= E[E[Y|X]] \= sum_x E[Y| X=x] cdot P{X=x}\=sum_x E[Y| X=x] cdot {1over 31}\=sum_{x=70}^{100} 0.75cdot x cdot {1over 31}\={3over 124}sum_{x=70}^{100}x\.\.\.\=63.75}$$
answered Mar 17 at 20:55
Mostafa AyazMostafa Ayaz
18.3k31040
$endgroup$
Your answer started true but continued wrong! Hence your answer, we can write:$$E[Y] {= E[E[Y|X]] \= sum_x E[Y| X=x] cdot P{X=x}\=sum_x E[Y| X=x] cdot {1over 31}\=sum_{x=70}^{100} 0.75cdot x cdot {1over 31}\={3over 124}sum_{x=70}^{100}x\.\.\.\=63.75}$$
answered Mar 17 at 20:55
Mostafa AyazMostafa Ayaz
18.3k31040
answered Mar 17 at 20:55
Mostafa AyazMostafa Ayaz
18.3k31040
answered Mar 17 at 20:55
Mostafa AyazMostafa Ayaz
18.3k31040
answered Mar 17 at 20:55
Mostafa AyazMostafa Ayaz
18.3k31040
18.3k31040
add a comment |
add a comment |
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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