Conditional expectation in Cambridge stats notes
0
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I am going over these, page 15.
They say that the optimal estimator is $$lambda^* = mathbb{E} left[ X_1 Bigg| sum_i X_i= tright]$$
I fail to understand how to evaluate this expectation. I also fail to understand why
$$mathbb{E}left[ sum_i X_i Bigg| sum_{i}X_i = tright] = t$$
TL;DR:
$X_1, ..., X_n sim P(lambda)$, they are IID and $T(X) = sum_i X_i = t$ is a sufficient statistic.
statistics
asked Dec 17 '18 at 15:51
i squared - Keep it Reali squared - Keep it Real
1,61311027
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add a comment |
0
$begingroup$
I am going over these, page 15.
They say that the optimal estimator is $$lambda^* = mathbb{E} left[ X_1 Bigg| sum_i X_i= tright]$$
I fail to understand how to evaluate this expectation. I also fail to understand why
$$mathbb{E}left[ sum_i X_i Bigg| sum_{i}X_i = tright] = t$$
TL;DR:
$X_1, ..., X_n sim P(lambda)$, they are IID and $T(X) = sum_i X_i = t$ is a sufficient statistic.
statistics
asked Dec 17 '18 at 15:51
i squared - Keep it Reali squared - Keep it Real
1,61311027
$endgroup$
$begingroup$
For your second expression, it is of the form $mathbb E[Y mid Y=y]$ which is clearly $y$
$endgroup$
– Henry
Dec 17 '18 at 16:01
$begingroup$
why? I do not know how to evaluate conditional expectation
$endgroup$
– i squared - Keep it Real
Dec 17 '18 at 16:03
$begingroup$
If you knew that $Y=y$ then the conditional expectation of $Y$ would be $y$. You can find out about conditional probability and conditional expectation in notes for the Probability Part Ia course, a precursor of these Statistics Part Ib course notes.
$endgroup$
– Henry
Dec 17 '18 at 16:10
$begingroup$
And the notes then tell you how to evaluate the first expression. Since the $X_i$ are independently and identically distributed and since expectation is linear, the second expression is $n$ times the first expression, making the first expression $dfrac{t}{n}$
$endgroup$
– Henry
Dec 17 '18 at 16:28
$begingroup$
@Henry ahh. I'll have a look at those. Thanks
$endgroup$
– i squared - Keep it Real
Dec 17 '18 at 16:35
add a comment |
0
0
0
$begingroup$
I am going over these, page 15.
They say that the optimal estimator is $$lambda^* = mathbb{E} left[ X_1 Bigg| sum_i X_i= tright]$$
I fail to understand how to evaluate this expectation. I also fail to understand why
$$mathbb{E}left[ sum_i X_i Bigg| sum_{i}X_i = tright] = t$$
TL;DR:
$X_1, ..., X_n sim P(lambda)$, they are IID and $T(X) = sum_i X_i = t$ is a sufficient statistic.
statistics
asked Dec 17 '18 at 15:51
i squared - Keep it Reali squared - Keep it Real
1,61311027
$endgroup$
I am going over these, page 15.
They say that the optimal estimator is $$lambda^* = mathbb{E} left[ X_1 Bigg| sum_i X_i= tright]$$
I fail to understand how to evaluate this expectation. I also fail to understand why
$$mathbb{E}left[ sum_i X_i Bigg| sum_{i}X_i = tright] = t$$
TL;DR:
$X_1, ..., X_n sim P(lambda)$, they are IID and $T(X) = sum_i X_i = t$ is a sufficient statistic.
statistics
statistics
asked Dec 17 '18 at 15:51
i squared - Keep it Reali squared - Keep it Real
1,61311027
asked Dec 17 '18 at 15:51
i squared - Keep it Reali squared - Keep it Real
1,61311027
asked Dec 17 '18 at 15:51
i squared - Keep it Reali squared - Keep it Real
1,61311027
asked Dec 17 '18 at 15:51
i squared - Keep it Reali squared - Keep it Real
1,61311027
asked Dec 17 '18 at 15:51
i squared - Keep it Reali squared - Keep it Real
1,61311027
1,61311027
$begingroup$
For your second expression, it is of the form $mathbb E[Y mid Y=y]$ which is clearly $y$
$endgroup$
– Henry
Dec 17 '18 at 16:01
$begingroup$
why? I do not know how to evaluate conditional expectation
$endgroup$
– i squared - Keep it Real
Dec 17 '18 at 16:03
$begingroup$
If you knew that $Y=y$ then the conditional expectation of $Y$ would be $y$. You can find out about conditional probability and conditional expectation in notes for the Probability Part Ia course, a precursor of these Statistics Part Ib course notes.
$endgroup$
– Henry
Dec 17 '18 at 16:10
$begingroup$
And the notes then tell you how to evaluate the first expression. Since the $X_i$ are independently and identically distributed and since expectation is linear, the second expression is $n$ times the first expression, making the first expression $dfrac{t}{n}$
$endgroup$
– Henry
Dec 17 '18 at 16:28
$begingroup$
@Henry ahh. I'll have a look at those. Thanks
$endgroup$
– i squared - Keep it Real
Dec 17 '18 at 16:35
add a comment |
$begingroup$
For your second expression, it is of the form $mathbb E[Y mid Y=y]$ which is clearly $y$
$endgroup$
– Henry
Dec 17 '18 at 16:01
$begingroup$
why? I do not know how to evaluate conditional expectation
$endgroup$
– i squared - Keep it Real
Dec 17 '18 at 16:03
$begingroup$
If you knew that $Y=y$ then the conditional expectation of $Y$ would be $y$. You can find out about conditional probability and conditional expectation in notes for the Probability Part Ia course, a precursor of these Statistics Part Ib course notes.
$endgroup$
– Henry
Dec 17 '18 at 16:10
$begingroup$
And the notes then tell you how to evaluate the first expression. Since the $X_i$ are independently and identically distributed and since expectation is linear, the second expression is $n$ times the first expression, making the first expression $dfrac{t}{n}$
$endgroup$
– Henry
Dec 17 '18 at 16:28
$begingroup$
@Henry ahh. I'll have a look at those. Thanks
$endgroup$
– i squared - Keep it Real
Dec 17 '18 at 16:35
$begingroup$
For your second expression, it is of the form $mathbb E[Y mid Y=y]$ which is clearly $y$
$endgroup$
– Henry
Dec 17 '18 at 16:01
$begingroup$
For your second expression, it is of the form $mathbb E[Y mid Y=y]$ which is clearly $y$
$endgroup$
– Henry
Dec 17 '18 at 16:01
$begingroup$
why? I do not know how to evaluate conditional expectation
$endgroup$
– i squared - Keep it Real
Dec 17 '18 at 16:03
$begingroup$
why? I do not know how to evaluate conditional expectation
$endgroup$
– i squared - Keep it Real
Dec 17 '18 at 16:03
$begingroup$
If you knew that $Y=y$ then the conditional expectation of $Y$ would be $y$. You can find out about conditional probability and conditional expectation in notes for the Probability Part Ia course, a precursor of these Statistics Part Ib course notes.
$endgroup$
– Henry
Dec 17 '18 at 16:10
$begingroup$
If you knew that $Y=y$ then the conditional expectation of $Y$ would be $y$. You can find out about conditional probability and conditional expectation in notes for the Probability Part Ia course, a precursor of these Statistics Part Ib course notes.
$endgroup$
– Henry
Dec 17 '18 at 16:10
$begingroup$
And the notes then tell you how to evaluate the first expression. Since the $X_i$ are independently and identically distributed and since expectation is linear, the second expression is $n$ times the first expression, making the first expression $dfrac{t}{n}$
$endgroup$
– Henry
Dec 17 '18 at 16:28
$begingroup$
And the notes then tell you how to evaluate the first expression. Since the $X_i$ are independently and identically distributed and since expectation is linear, the second expression is $n$ times the first expression, making the first expression $dfrac{t}{n}$
$endgroup$
– Henry
Dec 17 '18 at 16:28
$begingroup$
@Henry ahh. I'll have a look at those. Thanks
$endgroup$
– i squared - Keep it Real
Dec 17 '18 at 16:35
$begingroup$
@Henry ahh. I'll have a look at those. Thanks
$endgroup$
– i squared - Keep it Real
Dec 17 '18 at 16:35
add a comment |
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$begingroup$
For your second expression, it is of the form $mathbb E[Y mid Y=y]$ which is clearly $y$
$endgroup$
– Henry
Dec 17 '18 at 16:01
$begingroup$
why? I do not know how to evaluate conditional expectation
$endgroup$
– i squared - Keep it Real
Dec 17 '18 at 16:03
$begingroup$
If you knew that $Y=y$ then the conditional expectation of $Y$ would be $y$. You can find out about conditional probability and conditional expectation in notes for the Probability Part Ia course, a precursor of these Statistics Part Ib course notes.
$endgroup$
– Henry
Dec 17 '18 at 16:10
$begingroup$
And the notes then tell you how to evaluate the first expression. Since the $X_i$ are independently and identically distributed and since expectation is linear, the second expression is $n$ times the first expression, making the first expression $dfrac{t}{n}$
$endgroup$
– Henry
Dec 17 '18 at 16:28
$begingroup$
@Henry ahh. I'll have a look at those. Thanks
$endgroup$
– i squared - Keep it Real
Dec 17 '18 at 16:35