Expectation of sum of independent Poisson distributions
I have three independent Poisson distributions:
$X_1 sim mathcal{P}(15)$,
$X_2 sim mathcal{P}(21)$ &
$X_3 sim mathcal{P}(10)$.
I wish to find the Expectation and Variance of $X_1 + X_2 + X_3$.
For the expectation, my first intuition was to add up the $lambda$s i.e. $15+21+10$ since they are independent. Similar reasoning for the variance. However, I'm not sure if my reasoning is correct.
I would appreciate any help.
self-learning poisson-distribution expected-value
edited Nov 19 '18 at 3:06
Tianlalu
3,09621038
asked Nov 19 '18 at 2:26
vic12
254
add a comment |
I have three independent Poisson distributions:
$X_1 sim mathcal{P}(15)$,
$X_2 sim mathcal{P}(21)$ &
$X_3 sim mathcal{P}(10)$.
I wish to find the Expectation and Variance of $X_1 + X_2 + X_3$.
For the expectation, my first intuition was to add up the $lambda$s i.e. $15+21+10$ since they are independent. Similar reasoning for the variance. However, I'm not sure if my reasoning is correct.
I would appreciate any help.
self-learning poisson-distribution expected-value
edited Nov 19 '18 at 3:06
Tianlalu
3,09621038
asked Nov 19 '18 at 2:26
vic12
254
add a comment |
I have three independent Poisson distributions:
$X_1 sim mathcal{P}(15)$,
$X_2 sim mathcal{P}(21)$ &
$X_3 sim mathcal{P}(10)$.
I wish to find the Expectation and Variance of $X_1 + X_2 + X_3$.
For the expectation, my first intuition was to add up the $lambda$s i.e. $15+21+10$ since they are independent. Similar reasoning for the variance. However, I'm not sure if my reasoning is correct.
I would appreciate any help.
self-learning poisson-distribution expected-value
edited Nov 19 '18 at 3:06
Tianlalu
3,09621038
asked Nov 19 '18 at 2:26
vic12
254
I have three independent Poisson distributions:
$X_1 sim mathcal{P}(15)$,
$X_2 sim mathcal{P}(21)$ &
$X_3 sim mathcal{P}(10)$.
I wish to find the Expectation and Variance of $X_1 + X_2 + X_3$.
For the expectation, my first intuition was to add up the $lambda$s i.e. $15+21+10$ since they are independent. Similar reasoning for the variance. However, I'm not sure if my reasoning is correct.
I would appreciate any help.
self-learning poisson-distribution expected-value
self-learning poisson-distribution expected-value
edited Nov 19 '18 at 3:06
Tianlalu
3,09621038
asked Nov 19 '18 at 2:26
vic12
254
edited Nov 19 '18 at 3:06
Tianlalu
3,09621038
asked Nov 19 '18 at 2:26
vic12
254
edited Nov 19 '18 at 3:06
Tianlalu
3,09621038
edited Nov 19 '18 at 3:06
Tianlalu
3,09621038
edited Nov 19 '18 at 3:06
Tianlalu
3,09621038
3,09621038
asked Nov 19 '18 at 2:26
vic12
254
asked Nov 19 '18 at 2:26
vic12
254
asked Nov 19 '18 at 2:26
vic12
254
254
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
Your three random variables are Poisson distributed. The sum of these random variables will also be Poisson. So your intuition was correct about the expectation.
$X1 + X2 + X3 sim mathcal{P}(15 + 21 + 10)$
https://en.wikipedia.org/wiki/Poisson_distribution#Properties (see the fourth section under properties of the Poisson distribution)
Now that you know the distribution of your sum you can take it from here!
answered Nov 19 '18 at 2:36
KnowsNothing
355
add a comment |
$E(sum_{i=1}^n X_{i}) = sum_{i=1}^n E(X_{i})$
$V(sum_{i=1}^n X_{i}) = sum_{i=1}^n V(X_{i})$ since $COV(X_i,X_j)=0$ due to independence.
So you are correct
answered Nov 19 '18 at 2:37
pfmr1995
93
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
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active
oldest
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Your three random variables are Poisson distributed. The sum of these random variables will also be Poisson. So your intuition was correct about the expectation.
$X1 + X2 + X3 sim mathcal{P}(15 + 21 + 10)$
https://en.wikipedia.org/wiki/Poisson_distribution#Properties (see the fourth section under properties of the Poisson distribution)
Now that you know the distribution of your sum you can take it from here!
answered Nov 19 '18 at 2:36
KnowsNothing
355
add a comment |
Your three random variables are Poisson distributed. The sum of these random variables will also be Poisson. So your intuition was correct about the expectation.
$X1 + X2 + X3 sim mathcal{P}(15 + 21 + 10)$
https://en.wikipedia.org/wiki/Poisson_distribution#Properties (see the fourth section under properties of the Poisson distribution)
Now that you know the distribution of your sum you can take it from here!
answered Nov 19 '18 at 2:36
KnowsNothing
355
add a comment |
Your three random variables are Poisson distributed. The sum of these random variables will also be Poisson. So your intuition was correct about the expectation.
$X1 + X2 + X3 sim mathcal{P}(15 + 21 + 10)$
https://en.wikipedia.org/wiki/Poisson_distribution#Properties (see the fourth section under properties of the Poisson distribution)
Now that you know the distribution of your sum you can take it from here!
answered Nov 19 '18 at 2:36
KnowsNothing
355
Your three random variables are Poisson distributed. The sum of these random variables will also be Poisson. So your intuition was correct about the expectation.
$X1 + X2 + X3 sim mathcal{P}(15 + 21 + 10)$
https://en.wikipedia.org/wiki/Poisson_distribution#Properties (see the fourth section under properties of the Poisson distribution)
Now that you know the distribution of your sum you can take it from here!
answered Nov 19 '18 at 2:36
KnowsNothing
355
answered Nov 19 '18 at 2:36
KnowsNothing
355
answered Nov 19 '18 at 2:36
KnowsNothing
355
answered Nov 19 '18 at 2:36
KnowsNothing
355
355
add a comment |
add a comment |
$E(sum_{i=1}^n X_{i}) = sum_{i=1}^n E(X_{i})$
$V(sum_{i=1}^n X_{i}) = sum_{i=1}^n V(X_{i})$ since $COV(X_i,X_j)=0$ due to independence.
So you are correct
answered Nov 19 '18 at 2:37
pfmr1995
93
add a comment |
$E(sum_{i=1}^n X_{i}) = sum_{i=1}^n E(X_{i})$
$V(sum_{i=1}^n X_{i}) = sum_{i=1}^n V(X_{i})$ since $COV(X_i,X_j)=0$ due to independence.
So you are correct
answered Nov 19 '18 at 2:37
pfmr1995
93
add a comment |
$E(sum_{i=1}^n X_{i}) = sum_{i=1}^n E(X_{i})$
$V(sum_{i=1}^n X_{i}) = sum_{i=1}^n V(X_{i})$ since $COV(X_i,X_j)=0$ due to independence.
So you are correct
answered Nov 19 '18 at 2:37
pfmr1995
93
$E(sum_{i=1}^n X_{i}) = sum_{i=1}^n E(X_{i})$
$V(sum_{i=1}^n X_{i}) = sum_{i=1}^n V(X_{i})$ since $COV(X_i,X_j)=0$ due to independence.
So you are correct
answered Nov 19 '18 at 2:37
pfmr1995
93
answered Nov 19 '18 at 2:37
pfmr1995
93
answered Nov 19 '18 at 2:37
pfmr1995
93
answered Nov 19 '18 at 2:37
pfmr1995
93
93
add a comment |
add a comment |
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