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March 12th, 2017, 03:57 PM   #1
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Sigma rule question

Hi everyone
I was wondering why this equality is true considering that $d_i$ is not a constant

$\sum E(d_{i}u_{i})=\sum d_{i}E(u_{i})$

$\[d_{i}=(x_{i}-\bar{x})\]$


Thank you in advance

Last edited by dthiaw; March 12th, 2017 at 04:03 PM.
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March 12th, 2017, 10:09 PM   #2
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if $d_i$ is a constant then for any random variable $u_i$

$E[d_i u_i] = d_i E[u_i]$

It's not clear whether $d_i$ is a rv or not.
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March 13th, 2017, 10:03 AM   #3
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yeah but in this case di
is definitely not a constant. that is why am surprised

Last edited by dthiaw; March 13th, 2017 at 10:06 AM.
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March 13th, 2017, 10:16 AM   #4
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Quote:
Originally Posted by dthiaw View Post
yeah but in this case di
is definitely not a constant. that is why am surprised
but each individual $d_i$ is.

The summation notation allows for each $d_i$ to be different.
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March 13th, 2017, 10:22 AM   #5
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ok I see what you mean. Thanks a bunch once again
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