Can anyone explain how this summation was simplified/rearranged? I can't for the life of me follow the steps
$begingroup$
See attached image. Part of a question asking to calculate the marginal pmf when the joint pmf is known.
probability summation
edited Jan 13 at 16:58
Parcly Taxel
44.7k1376109
asked Jan 13 at 16:55
frenny13frenny13
12
$endgroup$
|
show 1 more comment
$begingroup$
See attached image. Part of a question asking to calculate the marginal pmf when the joint pmf is known.
probability summation
edited Jan 13 at 16:58
Parcly Taxel
44.7k1376109
asked Jan 13 at 16:55
frenny13frenny13
12
$endgroup$
1
$begingroup$
How far can you get? Which step do you not understand?
$endgroup$
– saulspatz
Jan 13 at 16:59
$begingroup$
The only thing I could think to take outside the summation was the e^(-lambda). How did the first summation become rearranged the second?
$endgroup$
– frenny13
Jan 13 at 17:05
$begingroup$
I don't understand what $sum$ means. Could you write out explicitly (as in $e^x = 1 + x + frac{x^2}{2!} + ldots $) what the first three or four terms in the expressions on the right are?
$endgroup$
– Dilip Sarwate
Jan 13 at 17:08
$begingroup$
Write out the binomial coefficient: $z!$ cancels in numerator and denominator. Also note that they have multiplied and divided by $lambda^y$ Note that $y$ is constant, so you can pull out $p^y$ for example.
$endgroup$
– saulspatz
Jan 13 at 17:09
$begingroup$
Ohh that makes a lot more sense forgot to write out the binomial coefficient now it all rearranges correctly thanks
$endgroup$
– frenny13
Jan 13 at 17:15
|
show 1 more comment
$begingroup$
See attached image. Part of a question asking to calculate the marginal pmf when the joint pmf is known.
probability summation
edited Jan 13 at 16:58
Parcly Taxel
44.7k1376109
asked Jan 13 at 16:55
frenny13frenny13
12
$endgroup$
See attached image. Part of a question asking to calculate the marginal pmf when the joint pmf is known.
probability summation
probability summation
edited Jan 13 at 16:58
Parcly Taxel
44.7k1376109
asked Jan 13 at 16:55
frenny13frenny13
12
edited Jan 13 at 16:58
Parcly Taxel
44.7k1376109
asked Jan 13 at 16:55
frenny13frenny13
12
edited Jan 13 at 16:58
Parcly Taxel
44.7k1376109
edited Jan 13 at 16:58
Parcly Taxel
44.7k1376109
edited Jan 13 at 16:58
Parcly Taxel
44.7k1376109
44.7k1376109
asked Jan 13 at 16:55
frenny13frenny13
12
asked Jan 13 at 16:55
frenny13frenny13
12
asked Jan 13 at 16:55
frenny13frenny13
12
12
1
$begingroup$
How far can you get? Which step do you not understand?
$endgroup$
– saulspatz
Jan 13 at 16:59
$begingroup$
The only thing I could think to take outside the summation was the e^(-lambda). How did the first summation become rearranged the second?
$endgroup$
– frenny13
Jan 13 at 17:05
$begingroup$
I don't understand what $sum$ means. Could you write out explicitly (as in $e^x = 1 + x + frac{x^2}{2!} + ldots $) what the first three or four terms in the expressions on the right are?
$endgroup$
– Dilip Sarwate
Jan 13 at 17:08
$begingroup$
Write out the binomial coefficient: $z!$ cancels in numerator and denominator. Also note that they have multiplied and divided by $lambda^y$ Note that $y$ is constant, so you can pull out $p^y$ for example.
$endgroup$
– saulspatz
Jan 13 at 17:09
$begingroup$
Ohh that makes a lot more sense forgot to write out the binomial coefficient now it all rearranges correctly thanks
$endgroup$
– frenny13
Jan 13 at 17:15
|
show 1 more comment
1
$begingroup$
How far can you get? Which step do you not understand?
$endgroup$
– saulspatz
Jan 13 at 16:59
$begingroup$
The only thing I could think to take outside the summation was the e^(-lambda). How did the first summation become rearranged the second?
$endgroup$
– frenny13
Jan 13 at 17:05
$begingroup$
I don't understand what $sum$ means. Could you write out explicitly (as in $e^x = 1 + x + frac{x^2}{2!} + ldots $) what the first three or four terms in the expressions on the right are?
$endgroup$
– Dilip Sarwate
Jan 13 at 17:08
$begingroup$
Write out the binomial coefficient: $z!$ cancels in numerator and denominator. Also note that they have multiplied and divided by $lambda^y$ Note that $y$ is constant, so you can pull out $p^y$ for example.
$endgroup$
– saulspatz
Jan 13 at 17:09
$begingroup$
Ohh that makes a lot more sense forgot to write out the binomial coefficient now it all rearranges correctly thanks
$endgroup$
– frenny13
Jan 13 at 17:15
1
1
$begingroup$
How far can you get? Which step do you not understand?
$endgroup$
– saulspatz
Jan 13 at 16:59
$begingroup$
How far can you get? Which step do you not understand?
$endgroup$
– saulspatz
Jan 13 at 16:59
$begingroup$
The only thing I could think to take outside the summation was the e^(-lambda). How did the first summation become rearranged the second?
$endgroup$
– frenny13
Jan 13 at 17:05
$begingroup$
The only thing I could think to take outside the summation was the e^(-lambda). How did the first summation become rearranged the second?
$endgroup$
– frenny13
Jan 13 at 17:05
$begingroup$
I don't understand what $sum$ means. Could you write out explicitly (as in $e^x = 1 + x + frac{x^2}{2!} + ldots $) what the first three or four terms in the expressions on the right are?
$endgroup$
– Dilip Sarwate
Jan 13 at 17:08
$begingroup$
I don't understand what $sum$ means. Could you write out explicitly (as in $e^x = 1 + x + frac{x^2}{2!} + ldots $) what the first three or four terms in the expressions on the right are?
$endgroup$
– Dilip Sarwate
Jan 13 at 17:08
$begingroup$
Write out the binomial coefficient: $z!$ cancels in numerator and denominator. Also note that they have multiplied and divided by $lambda^y$ Note that $y$ is constant, so you can pull out $p^y$ for example.
$endgroup$
– saulspatz
Jan 13 at 17:09
$begingroup$
Write out the binomial coefficient: $z!$ cancels in numerator and denominator. Also note that they have multiplied and divided by $lambda^y$ Note that $y$ is constant, so you can pull out $p^y$ for example.
$endgroup$
– saulspatz
Jan 13 at 17:09
$begingroup$
Ohh that makes a lot more sense forgot to write out the binomial coefficient now it all rearranges correctly thanks
$endgroup$
– frenny13
Jan 13 at 17:15
$begingroup$
Ohh that makes a lot more sense forgot to write out the binomial coefficient now it all rearranges correctly thanks
$endgroup$
– frenny13
Jan 13 at 17:15
|
show 1 more comment
2 Answers
2
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oldest
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$begingroup$
$P(Y=y)=sum_{z=y}^inftybinom{z}{y}p^y(1-p)^{z-y}frac{e^{-lambda}lambda ^z}{z!}$.We have $
binom{z}{y}=frac{z!}{y!(z-y)!}$ and $lambda^z=lambda^ylambda^{z-y}$ givimg second term.
Let $k=z-y$ so the second term sum becomes $sum_{k=0}^infty frac{(lambda(1-p))^k}{k!}=e^{lambda(1-p)}$ giving the third term.
answered Jan 13 at 17:15
herb steinbergherb steinberg
3,0382310
$endgroup$
add a comment |
$begingroup$
In the first one the summand goes like this$$binom{z}{y}p^y(1-p)^{z-y}{e^{-lambda}lambda^zover z!}{={z!over y!(z-y!)}p^y(1-p)^{z-y}{e^{-lambda}lambda^zover z!}\={1over y!(z-y!)}p^y(1-p)^{z-y}{e^{-lambda}lambda^zover 1}\={e^{-lambda}(plambda)^yover y!}{(1-p)^{z-y}lambda^{z-y}over(z-y)!}}$$after all, substitute $u=z-y$ and conclude.
answered Jan 13 at 17:17
Mostafa AyazMostafa Ayaz
16.6k3939
$endgroup$
add a comment |
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2 Answers
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$begingroup$
$P(Y=y)=sum_{z=y}^inftybinom{z}{y}p^y(1-p)^{z-y}frac{e^{-lambda}lambda ^z}{z!}$.We have $
binom{z}{y}=frac{z!}{y!(z-y)!}$ and $lambda^z=lambda^ylambda^{z-y}$ givimg second term.
Let $k=z-y$ so the second term sum becomes $sum_{k=0}^infty frac{(lambda(1-p))^k}{k!}=e^{lambda(1-p)}$ giving the third term.
answered Jan 13 at 17:15
herb steinbergherb steinberg
3,0382310
$endgroup$
add a comment |
$begingroup$
$P(Y=y)=sum_{z=y}^inftybinom{z}{y}p^y(1-p)^{z-y}frac{e^{-lambda}lambda ^z}{z!}$.We have $
binom{z}{y}=frac{z!}{y!(z-y)!}$ and $lambda^z=lambda^ylambda^{z-y}$ givimg second term.
Let $k=z-y$ so the second term sum becomes $sum_{k=0}^infty frac{(lambda(1-p))^k}{k!}=e^{lambda(1-p)}$ giving the third term.
answered Jan 13 at 17:15
herb steinbergherb steinberg
3,0382310
$endgroup$
add a comment |
$begingroup$
$P(Y=y)=sum_{z=y}^inftybinom{z}{y}p^y(1-p)^{z-y}frac{e^{-lambda}lambda ^z}{z!}$.We have $
binom{z}{y}=frac{z!}{y!(z-y)!}$ and $lambda^z=lambda^ylambda^{z-y}$ givimg second term.
Let $k=z-y$ so the second term sum becomes $sum_{k=0}^infty frac{(lambda(1-p))^k}{k!}=e^{lambda(1-p)}$ giving the third term.
answered Jan 13 at 17:15
herb steinbergherb steinberg
3,0382310
$endgroup$
$P(Y=y)=sum_{z=y}^inftybinom{z}{y}p^y(1-p)^{z-y}frac{e^{-lambda}lambda ^z}{z!}$.We have $
binom{z}{y}=frac{z!}{y!(z-y)!}$ and $lambda^z=lambda^ylambda^{z-y}$ givimg second term.
Let $k=z-y$ so the second term sum becomes $sum_{k=0}^infty frac{(lambda(1-p))^k}{k!}=e^{lambda(1-p)}$ giving the third term.
answered Jan 13 at 17:15
herb steinbergherb steinberg
3,0382310
answered Jan 13 at 17:15
herb steinbergherb steinberg
3,0382310
answered Jan 13 at 17:15
herb steinbergherb steinberg
3,0382310
answered Jan 13 at 17:15
herb steinbergherb steinberg
3,0382310
3,0382310
add a comment |
add a comment |
$begingroup$
In the first one the summand goes like this$$binom{z}{y}p^y(1-p)^{z-y}{e^{-lambda}lambda^zover z!}{={z!over y!(z-y!)}p^y(1-p)^{z-y}{e^{-lambda}lambda^zover z!}\={1over y!(z-y!)}p^y(1-p)^{z-y}{e^{-lambda}lambda^zover 1}\={e^{-lambda}(plambda)^yover y!}{(1-p)^{z-y}lambda^{z-y}over(z-y)!}}$$after all, substitute $u=z-y$ and conclude.
answered Jan 13 at 17:17
Mostafa AyazMostafa Ayaz
16.6k3939
$endgroup$
add a comment |
$begingroup$
In the first one the summand goes like this$$binom{z}{y}p^y(1-p)^{z-y}{e^{-lambda}lambda^zover z!}{={z!over y!(z-y!)}p^y(1-p)^{z-y}{e^{-lambda}lambda^zover z!}\={1over y!(z-y!)}p^y(1-p)^{z-y}{e^{-lambda}lambda^zover 1}\={e^{-lambda}(plambda)^yover y!}{(1-p)^{z-y}lambda^{z-y}over(z-y)!}}$$after all, substitute $u=z-y$ and conclude.
answered Jan 13 at 17:17
Mostafa AyazMostafa Ayaz
16.6k3939
$endgroup$
add a comment |
$begingroup$
In the first one the summand goes like this$$binom{z}{y}p^y(1-p)^{z-y}{e^{-lambda}lambda^zover z!}{={z!over y!(z-y!)}p^y(1-p)^{z-y}{e^{-lambda}lambda^zover z!}\={1over y!(z-y!)}p^y(1-p)^{z-y}{e^{-lambda}lambda^zover 1}\={e^{-lambda}(plambda)^yover y!}{(1-p)^{z-y}lambda^{z-y}over(z-y)!}}$$after all, substitute $u=z-y$ and conclude.
answered Jan 13 at 17:17
Mostafa AyazMostafa Ayaz
16.6k3939
$endgroup$
In the first one the summand goes like this$$binom{z}{y}p^y(1-p)^{z-y}{e^{-lambda}lambda^zover z!}{={z!over y!(z-y!)}p^y(1-p)^{z-y}{e^{-lambda}lambda^zover z!}\={1over y!(z-y!)}p^y(1-p)^{z-y}{e^{-lambda}lambda^zover 1}\={e^{-lambda}(plambda)^yover y!}{(1-p)^{z-y}lambda^{z-y}over(z-y)!}}$$after all, substitute $u=z-y$ and conclude.
answered Jan 13 at 17:17
Mostafa AyazMostafa Ayaz
16.6k3939
answered Jan 13 at 17:17
Mostafa AyazMostafa Ayaz
16.6k3939
answered Jan 13 at 17:17
Mostafa AyazMostafa Ayaz
16.6k3939
answered Jan 13 at 17:17
Mostafa AyazMostafa Ayaz
16.6k3939
16.6k3939
add a comment |
add a comment |
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1
$begingroup$
How far can you get? Which step do you not understand?
$endgroup$
– saulspatz
Jan 13 at 16:59
$begingroup$
The only thing I could think to take outside the summation was the e^(-lambda). How did the first summation become rearranged the second?
$endgroup$
– frenny13
Jan 13 at 17:05
$begingroup$
I don't understand what $sum$ means. Could you write out explicitly (as in $e^x = 1 + x + frac{x^2}{2!} + ldots $) what the first three or four terms in the expressions on the right are?
$endgroup$
– Dilip Sarwate
Jan 13 at 17:08
$begingroup$
Write out the binomial coefficient: $z!$ cancels in numerator and denominator. Also note that they have multiplied and divided by $lambda^y$ Note that $y$ is constant, so you can pull out $p^y$ for example.
$endgroup$
– saulspatz
Jan 13 at 17:09
$begingroup$
Ohh that makes a lot more sense forgot to write out the binomial coefficient now it all rearranges correctly thanks
$endgroup$
– frenny13
Jan 13 at 17:15