Simplification of Summation for an Expected Value Problem
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I am not good with series or summation notation. Could someone explain how they simplified the expression in this image. Thanks!
summation random-variables expected-value
asked Dec 5 '18 at 23:19
Raoul DukeRaoul Duke
34
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add a comment |
$begingroup$
I am not good with series or summation notation. Could someone explain how they simplified the expression in this image. Thanks!
summation random-variables expected-value
asked Dec 5 '18 at 23:19
Raoul DukeRaoul Duke
34
$endgroup$
add a comment |
$begingroup$
I am not good with series or summation notation. Could someone explain how they simplified the expression in this image. Thanks!
summation random-variables expected-value
asked Dec 5 '18 at 23:19
Raoul DukeRaoul Duke
34
$endgroup$
I am not good with series or summation notation. Could someone explain how they simplified the expression in this image. Thanks!
summation random-variables expected-value
summation random-variables expected-value
asked Dec 5 '18 at 23:19
Raoul DukeRaoul Duke
34
asked Dec 5 '18 at 23:19
Raoul DukeRaoul Duke
34
asked Dec 5 '18 at 23:19
Raoul DukeRaoul Duke
34
asked Dec 5 '18 at 23:19
Raoul DukeRaoul Duke
34
asked Dec 5 '18 at 23:19
Raoul DukeRaoul Duke
34
34
add a comment |
add a comment |
1 Answer
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Since $k=0$ and $k=1$, the sum becomes
$$sum_{k=0}^1sum_{ell = 0}^3 3(1+k)ell p_{X,Y}(k,ell) = sum_{ell = 0}^3 left[3 ell p_{X,Y}(0,ell) + 3(1+1)ell p_{X,Y}(1,ell)right]$$
$$=$$
$$sum_{ell = 0}^3 3ell p_{X,Y}(0,ell) + sum_{ell = 0}^36ell p_{X,Y}(1,ell)$$
simply by using the fact that :
$$sum(a+b) = sum a + sum b$$
answered Dec 5 '18 at 23:25
RebellosRebellos
14.5k31246
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$begingroup$
Thank you. I think the variables confused me
$endgroup$
– Raoul Duke
Dec 5 '18 at 23:30
$begingroup$
No problem, if you are a begginer, summations usually scare people away ! Glad I could help.
$endgroup$
– Rebellos
Dec 5 '18 at 23:31
$begingroup$
Is there a reason I got down-voted by someone? Not upset, just want to make sure I'm following the community guidelines correctly.
$endgroup$
– Raoul Duke
Dec 6 '18 at 0:14
add a comment |
Your Answer
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1 Answer
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1 Answer
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oldest
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$begingroup$
Since $k=0$ and $k=1$, the sum becomes
$$sum_{k=0}^1sum_{ell = 0}^3 3(1+k)ell p_{X,Y}(k,ell) = sum_{ell = 0}^3 left[3 ell p_{X,Y}(0,ell) + 3(1+1)ell p_{X,Y}(1,ell)right]$$
$$=$$
$$sum_{ell = 0}^3 3ell p_{X,Y}(0,ell) + sum_{ell = 0}^36ell p_{X,Y}(1,ell)$$
simply by using the fact that :
$$sum(a+b) = sum a + sum b$$
answered Dec 5 '18 at 23:25
RebellosRebellos
14.5k31246
$endgroup$
$begingroup$
Thank you. I think the variables confused me
$endgroup$
– Raoul Duke
Dec 5 '18 at 23:30
$begingroup$
No problem, if you are a begginer, summations usually scare people away ! Glad I could help.
$endgroup$
– Rebellos
Dec 5 '18 at 23:31
$begingroup$
Is there a reason I got down-voted by someone? Not upset, just want to make sure I'm following the community guidelines correctly.
$endgroup$
– Raoul Duke
Dec 6 '18 at 0:14
add a comment |
$begingroup$
Since $k=0$ and $k=1$, the sum becomes
$$sum_{k=0}^1sum_{ell = 0}^3 3(1+k)ell p_{X,Y}(k,ell) = sum_{ell = 0}^3 left[3 ell p_{X,Y}(0,ell) + 3(1+1)ell p_{X,Y}(1,ell)right]$$
$$=$$
$$sum_{ell = 0}^3 3ell p_{X,Y}(0,ell) + sum_{ell = 0}^36ell p_{X,Y}(1,ell)$$
simply by using the fact that :
$$sum(a+b) = sum a + sum b$$
answered Dec 5 '18 at 23:25
RebellosRebellos
14.5k31246
$endgroup$
$begingroup$
Thank you. I think the variables confused me
$endgroup$
– Raoul Duke
Dec 5 '18 at 23:30
$begingroup$
No problem, if you are a begginer, summations usually scare people away ! Glad I could help.
$endgroup$
– Rebellos
Dec 5 '18 at 23:31
$begingroup$
Is there a reason I got down-voted by someone? Not upset, just want to make sure I'm following the community guidelines correctly.
$endgroup$
– Raoul Duke
Dec 6 '18 at 0:14
add a comment |
$begingroup$
Since $k=0$ and $k=1$, the sum becomes
$$sum_{k=0}^1sum_{ell = 0}^3 3(1+k)ell p_{X,Y}(k,ell) = sum_{ell = 0}^3 left[3 ell p_{X,Y}(0,ell) + 3(1+1)ell p_{X,Y}(1,ell)right]$$
$$=$$
$$sum_{ell = 0}^3 3ell p_{X,Y}(0,ell) + sum_{ell = 0}^36ell p_{X,Y}(1,ell)$$
simply by using the fact that :
$$sum(a+b) = sum a + sum b$$
answered Dec 5 '18 at 23:25
RebellosRebellos
14.5k31246
$endgroup$
Since $k=0$ and $k=1$, the sum becomes
$$sum_{k=0}^1sum_{ell = 0}^3 3(1+k)ell p_{X,Y}(k,ell) = sum_{ell = 0}^3 left[3 ell p_{X,Y}(0,ell) + 3(1+1)ell p_{X,Y}(1,ell)right]$$
$$=$$
$$sum_{ell = 0}^3 3ell p_{X,Y}(0,ell) + sum_{ell = 0}^36ell p_{X,Y}(1,ell)$$
simply by using the fact that :
$$sum(a+b) = sum a + sum b$$
answered Dec 5 '18 at 23:25
RebellosRebellos
14.5k31246
answered Dec 5 '18 at 23:25
RebellosRebellos
14.5k31246
answered Dec 5 '18 at 23:25
RebellosRebellos
14.5k31246
answered Dec 5 '18 at 23:25
RebellosRebellos
14.5k31246
14.5k31246
$begingroup$
Thank you. I think the variables confused me
$endgroup$
– Raoul Duke
Dec 5 '18 at 23:30
$begingroup$
No problem, if you are a begginer, summations usually scare people away ! Glad I could help.
$endgroup$
– Rebellos
Dec 5 '18 at 23:31
$begingroup$
Is there a reason I got down-voted by someone? Not upset, just want to make sure I'm following the community guidelines correctly.
$endgroup$
– Raoul Duke
Dec 6 '18 at 0:14
add a comment |
$begingroup$
Thank you. I think the variables confused me
$endgroup$
– Raoul Duke
Dec 5 '18 at 23:30
$begingroup$
No problem, if you are a begginer, summations usually scare people away ! Glad I could help.
$endgroup$
– Rebellos
Dec 5 '18 at 23:31
$begingroup$
Is there a reason I got down-voted by someone? Not upset, just want to make sure I'm following the community guidelines correctly.
$endgroup$
– Raoul Duke
Dec 6 '18 at 0:14
$begingroup$
Thank you. I think the variables confused me
$endgroup$
– Raoul Duke
Dec 5 '18 at 23:30
$begingroup$
Thank you. I think the variables confused me
$endgroup$
– Raoul Duke
Dec 5 '18 at 23:30
$begingroup$
No problem, if you are a begginer, summations usually scare people away ! Glad I could help.
$endgroup$
– Rebellos
Dec 5 '18 at 23:31
$begingroup$
No problem, if you are a begginer, summations usually scare people away ! Glad I could help.
$endgroup$
– Rebellos
Dec 5 '18 at 23:31
$begingroup$
Is there a reason I got down-voted by someone? Not upset, just want to make sure I'm following the community guidelines correctly.
$endgroup$
– Raoul Duke
Dec 6 '18 at 0:14
$begingroup$
Is there a reason I got down-voted by someone? Not upset, just want to make sure I'm following the community guidelines correctly.
$endgroup$
– Raoul Duke
Dec 6 '18 at 0:14
add a comment |
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