Expectation of sum of independent Poisson distributions
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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 at 3:06
Tianlalu
2,9901936
asked Nov 19 at 2:26
vic12
254
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up vote
0
down vote
favorite
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 at 3:06
Tianlalu
2,9901936
asked Nov 19 at 2:26
vic12
254
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
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 at 3:06
Tianlalu
2,9901936
asked Nov 19 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 at 3:06
Tianlalu
2,9901936
asked Nov 19 at 2:26
vic12
254
edited Nov 19 at 3:06
Tianlalu
2,9901936
asked Nov 19 at 2:26
vic12
254
edited Nov 19 at 3:06
Tianlalu
2,9901936
edited Nov 19 at 3:06
Tianlalu
2,9901936
edited Nov 19 at 3:06
Tianlalu
2,9901936
2,9901936
asked Nov 19 at 2:26
vic12
254
asked Nov 19 at 2:26
vic12
254
asked Nov 19 at 2:26
vic12
254
254
add a comment |
add a comment |
2 Answers
2
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oldest
votes
up vote
0
down vote
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 at 2:36
KnowsNothing
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$E(sum_{i=1}^n X_{i}) = sum_{i=1}^n E(X_{i})$
$V(sum_{i=1}^n X_{i}) = sum_{i=1}^n V(X_{i})$ since $COV(X_i,X_j)=0$ due to independence.
So you are correct
answered Nov 19 at 2:37
pfmr1995
113
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
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 at 2:36
KnowsNothing
355
add a comment |
up vote
0
down vote
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 at 2:36
KnowsNothing
355
add a comment |
up vote
0
down vote
up vote
0
down vote
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 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 at 2:36
KnowsNothing
355
answered Nov 19 at 2:36
KnowsNothing
355
answered Nov 19 at 2:36
KnowsNothing
355
answered Nov 19 at 2:36
KnowsNothing
355
355
add a comment |
add a comment |
up vote
0
down vote
$E(sum_{i=1}^n X_{i}) = sum_{i=1}^n E(X_{i})$
$V(sum_{i=1}^n X_{i}) = sum_{i=1}^n V(X_{i})$ since $COV(X_i,X_j)=0$ due to independence.
So you are correct
answered Nov 19 at 2:37
pfmr1995
113
add a comment |
up vote
0
down vote
$E(sum_{i=1}^n X_{i}) = sum_{i=1}^n E(X_{i})$
$V(sum_{i=1}^n X_{i}) = sum_{i=1}^n V(X_{i})$ since $COV(X_i,X_j)=0$ due to independence.
So you are correct
answered Nov 19 at 2:37
pfmr1995
113
add a comment |
up vote
0
down vote
up vote
0
down vote
$E(sum_{i=1}^n X_{i}) = sum_{i=1}^n E(X_{i})$
$V(sum_{i=1}^n X_{i}) = sum_{i=1}^n V(X_{i})$ since $COV(X_i,X_j)=0$ due to independence.
So you are correct
answered Nov 19 at 2:37
pfmr1995
113
$E(sum_{i=1}^n X_{i}) = sum_{i=1}^n E(X_{i})$
$V(sum_{i=1}^n X_{i}) = sum_{i=1}^n V(X_{i})$ since $COV(X_i,X_j)=0$ due to independence.
So you are correct
answered Nov 19 at 2:37
pfmr1995
113
answered Nov 19 at 2:37
pfmr1995
113
answered Nov 19 at 2:37
pfmr1995
113
answered Nov 19 at 2:37
pfmr1995
113
113
add a comment |
add a comment |
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