Mixing
Expected value of Bernoulli variable (for Linear regression model)
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Can anybody explain how the following reduction happens:
statistics expected-value
asked Jan 13 at 22:38
AmirAmir
1034
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add a comment |
$begingroup$
Can anybody explain how the following reduction happens:
statistics expected-value
asked Jan 13 at 22:38
AmirAmir
1034
$endgroup$
add a comment |
$begingroup$
Can anybody explain how the following reduction happens:
statistics expected-value
asked Jan 13 at 22:38
AmirAmir
1034
$endgroup$
Can anybody explain how the following reduction happens:
statistics expected-value
statistics expected-value
asked Jan 13 at 22:38
AmirAmir
1034
asked Jan 13 at 22:38
AmirAmir
1034
edited Jan 13 at 22:51
Amir
asked Jan 13 at 22:38
AmirAmir
1034
asked Jan 13 at 22:38
AmirAmir
1034
asked Jan 13 at 22:38
AmirAmir
1034
1034
add a comment |
add a comment |
1 Answer
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$begingroup$
begin{align}
mathbb{E} |y - (R*X) w|^2
&= p |y - X w|^2 + (1-p) |y|^2
\
&= p(|y|^2 - 2 y^top X w + |Xw|^2) + (1-p) |y|^2
\
&= (|y|^2 - 2 y^top (p X w) + p^2 |Xw|^2) - p^2 |Xw|^2 + p |Xw|^2
\
&= |y - pXw|^2 + p(1-p) |Xw|^2.
end{align}
answered Jan 13 at 22:43
angryavianangryavian
41.1k23380
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$begingroup$
Thank you so much, it was very helpful. Would you please give me a clue what is happened in the first line? Is it Bernoulli property?
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– Amir
Jan 13 at 23:03
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@Amir I used the law of total expectation with the two cases $R=0$ and $R=1$.
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– angryavian
Jan 13 at 23:09
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Thanks for your help.
$endgroup$
– Amir
Jan 13 at 23:22
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
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active
oldest
votes
$begingroup$
begin{align}
mathbb{E} |y - (R*X) w|^2
&= p |y - X w|^2 + (1-p) |y|^2
\
&= p(|y|^2 - 2 y^top X w + |Xw|^2) + (1-p) |y|^2
\
&= (|y|^2 - 2 y^top (p X w) + p^2 |Xw|^2) - p^2 |Xw|^2 + p |Xw|^2
\
&= |y - pXw|^2 + p(1-p) |Xw|^2.
end{align}
answered Jan 13 at 22:43
angryavianangryavian
41.1k23380
$endgroup$
$begingroup$
Thank you so much, it was very helpful. Would you please give me a clue what is happened in the first line? Is it Bernoulli property?
$endgroup$
– Amir
Jan 13 at 23:03
$begingroup$
@Amir I used the law of total expectation with the two cases $R=0$ and $R=1$.
$endgroup$
– angryavian
Jan 13 at 23:09
$begingroup$
Thanks for your help.
$endgroup$
– Amir
Jan 13 at 23:22
add a comment |
$begingroup$
begin{align}
mathbb{E} |y - (R*X) w|^2
&= p |y - X w|^2 + (1-p) |y|^2
\
&= p(|y|^2 - 2 y^top X w + |Xw|^2) + (1-p) |y|^2
\
&= (|y|^2 - 2 y^top (p X w) + p^2 |Xw|^2) - p^2 |Xw|^2 + p |Xw|^2
\
&= |y - pXw|^2 + p(1-p) |Xw|^2.
end{align}
answered Jan 13 at 22:43
angryavianangryavian
41.1k23380
$endgroup$
$begingroup$
Thank you so much, it was very helpful. Would you please give me a clue what is happened in the first line? Is it Bernoulli property?
$endgroup$
– Amir
Jan 13 at 23:03
$begingroup$
@Amir I used the law of total expectation with the two cases $R=0$ and $R=1$.
$endgroup$
– angryavian
Jan 13 at 23:09
$begingroup$
Thanks for your help.
$endgroup$
– Amir
Jan 13 at 23:22
add a comment |
$begingroup$
begin{align}
mathbb{E} |y - (R*X) w|^2
&= p |y - X w|^2 + (1-p) |y|^2
\
&= p(|y|^2 - 2 y^top X w + |Xw|^2) + (1-p) |y|^2
\
&= (|y|^2 - 2 y^top (p X w) + p^2 |Xw|^2) - p^2 |Xw|^2 + p |Xw|^2
\
&= |y - pXw|^2 + p(1-p) |Xw|^2.
end{align}
answered Jan 13 at 22:43
angryavianangryavian
41.1k23380
$endgroup$
begin{align}
mathbb{E} |y - (R*X) w|^2
&= p |y - X w|^2 + (1-p) |y|^2
\
&= p(|y|^2 - 2 y^top X w + |Xw|^2) + (1-p) |y|^2
\
&= (|y|^2 - 2 y^top (p X w) + p^2 |Xw|^2) - p^2 |Xw|^2 + p |Xw|^2
\
&= |y - pXw|^2 + p(1-p) |Xw|^2.
end{align}
answered Jan 13 at 22:43
angryavianangryavian
41.1k23380
answered Jan 13 at 22:43
angryavianangryavian
41.1k23380
answered Jan 13 at 22:43
angryavianangryavian
41.1k23380
answered Jan 13 at 22:43
angryavianangryavian
41.1k23380
41.1k23380
$begingroup$
Thank you so much, it was very helpful. Would you please give me a clue what is happened in the first line? Is it Bernoulli property?
$endgroup$
– Amir
Jan 13 at 23:03
$begingroup$
@Amir I used the law of total expectation with the two cases $R=0$ and $R=1$.
$endgroup$
– angryavian
Jan 13 at 23:09
$begingroup$
Thanks for your help.
$endgroup$
– Amir
Jan 13 at 23:22
add a comment |
$begingroup$
Thank you so much, it was very helpful. Would you please give me a clue what is happened in the first line? Is it Bernoulli property?
$endgroup$
– Amir
Jan 13 at 23:03
$begingroup$
@Amir I used the law of total expectation with the two cases $R=0$ and $R=1$.
$endgroup$
– angryavian
Jan 13 at 23:09
$begingroup$
Thanks for your help.
$endgroup$
– Amir
Jan 13 at 23:22
$begingroup$
Thank you so much, it was very helpful. Would you please give me a clue what is happened in the first line? Is it Bernoulli property?
$endgroup$
– Amir
Jan 13 at 23:03
$begingroup$
Thank you so much, it was very helpful. Would you please give me a clue what is happened in the first line? Is it Bernoulli property?
$endgroup$
– Amir
Jan 13 at 23:03
$begingroup$
@Amir I used the law of total expectation with the two cases $R=0$ and $R=1$.
$endgroup$
– angryavian
Jan 13 at 23:09
$begingroup$
@Amir I used the law of total expectation with the two cases $R=0$ and $R=1$.
$endgroup$
– angryavian
Jan 13 at 23:09
$begingroup$
Thanks for your help.
$endgroup$
– Amir
Jan 13 at 23:22
$begingroup$
Thanks for your help.
$endgroup$
– Amir
Jan 13 at 23:22
add a comment |
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