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Poisson distribution with independent, identically distributed $X_i,$ )

Cumulative distribution function or density for Compound Poisson distributionApproximate as Independent Identically distributedDeriving the distribution of poisson random variables.Determine the distribution of the sum of n independent identically distrubted poisson random variable $X_i$?Joint Distribution of n Poisson Random Variablesfunctions of identically distributed variables are identically distributedIndependent and identically distribued random variables $X_i$ with $X_1 sim exp(1/2)$independent Poisson-distribution random variableLimit of $S_N = sum_i=0^N X_i$ where $X_i$ is Laplace distributed and $N$ is Poisson distributed.Sum of Independent Poisson Distribution

0

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How can I find E($tildeX$) and Var($tildeX$) of a Poisson distribution with independent, identically distributed $X_i,$ )

probability poisson-distribution

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edited Mar 31 at 1:42

Nathan

asked Mar 29 at 3:22

NathanNathan

54

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0

$begingroup$

How can I find E($tildeX$) and Var($tildeX$) of a Poisson distribution with independent, identically distributed $X_i,$ )

probability poisson-distribution

share|cite|improve this question

edited Mar 31 at 1:42

Nathan

asked Mar 29 at 3:22

NathanNathan

54

$endgroup$

add a comment | 

$begingroup$

How can I find E($tildeX$) and Var($tildeX$) of a Poisson distribution with independent, identically distributed $X_i,$ )

probability poisson-distribution

share|cite|improve this question

edited Mar 31 at 1:42

Nathan

asked Mar 29 at 3:22

NathanNathan

54

$endgroup$

share|cite|improve this question

edited Mar 31 at 1:42

Nathan

asked Mar 29 at 3:22

NathanNathan

54

share|cite|improve this question

edited Mar 31 at 1:42

Nathan

asked Mar 29 at 3:22

NathanNathan

54

asked Mar 29 at 3:22

NathanNathan

54

asked Mar 29 at 3:22

NathanNathan

54

2 Answers
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You still need the sum signs in the last step there; it should be $frac1nleft(sum_i=1^n E(X_i)right)$ and $frac1n^2left(sum_i=1^n V(X_i)right)$.

How can I finish it for $X_i$ iid Poisson($lambda$)?

What is the mean of a Poisson distribution with parameter $lambda$? The variance? Then we add up $n$ copies of that, and divide by $n$ or $n^2$.

share|cite|improve this answer

answered Mar 29 at 3:38

jmerryjmerry

17k11633

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0

$begingroup$

I assume $tilde X$ is the same as $bar X$...

a) You are almost there but $E sum_i x_i = n E X$ given $X$ is i.i.d, hence the first answer $E bar X = E X$.

b) ditto here: $V sum_i X_i = n V X$ if $X$ is i.i.d., and hence the answer is $frac1n V X$.

Note that it depends on the i.i.d assumption, but not on the Poisson distribiton.

share|cite|improve this answer

answered Mar 29 at 3:41

Ott ToometOtt Toomet

1114

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0

$begingroup$

You still need the sum signs in the last step there; it should be $frac1nleft(sum_i=1^n E(X_i)right)$ and $frac1n^2left(sum_i=1^n V(X_i)right)$.

How can I finish it for $X_i$ iid Poisson($lambda$)?

What is the mean of a Poisson distribution with parameter $lambda$? The variance? Then we add up $n$ copies of that, and divide by $n$ or $n^2$.

share|cite|improve this answer

answered Mar 29 at 3:38

jmerryjmerry

17k11633

$endgroup$

add a comment | 

0

$begingroup$

You still need the sum signs in the last step there; it should be $frac1nleft(sum_i=1^n E(X_i)right)$ and $frac1n^2left(sum_i=1^n V(X_i)right)$.

How can I finish it for $X_i$ iid Poisson($lambda$)?

What is the mean of a Poisson distribution with parameter $lambda$? The variance? Then we add up $n$ copies of that, and divide by $n$ or $n^2$.

share|cite|improve this answer

answered Mar 29 at 3:38

jmerryjmerry

17k11633

$endgroup$

add a comment | 

$begingroup$

You still need the sum signs in the last step there; it should be $frac1nleft(sum_i=1^n E(X_i)right)$ and $frac1n^2left(sum_i=1^n V(X_i)right)$.

How can I finish it for $X_i$ iid Poisson($lambda$)?

What is the mean of a Poisson distribution with parameter $lambda$? The variance? Then we add up $n$ copies of that, and divide by $n$ or $n^2$.

share|cite|improve this answer

answered Mar 29 at 3:38

jmerryjmerry

17k11633

$endgroup$

share|cite|improve this answer

answered Mar 29 at 3:38

jmerryjmerry

17k11633

answered Mar 29 at 3:38

jmerryjmerry

17k11633

answered Mar 29 at 3:38

jmerryjmerry

17k11633

0

$begingroup$

I assume $tilde X$ is the same as $bar X$...

a) You are almost there but $E sum_i x_i = n E X$ given $X$ is i.i.d, hence the first answer $E bar X = E X$.

b) ditto here: $V sum_i X_i = n V X$ if $X$ is i.i.d., and hence the answer is $frac1n V X$.

Note that it depends on the i.i.d assumption, but not on the Poisson distribiton.

share|cite|improve this answer

answered Mar 29 at 3:41

Ott ToometOtt Toomet

1114

$endgroup$

add a comment | 

0

$begingroup$

I assume $tilde X$ is the same as $bar X$...

a) You are almost there but $E sum_i x_i = n E X$ given $X$ is i.i.d, hence the first answer $E bar X = E X$.

b) ditto here: $V sum_i X_i = n V X$ if $X$ is i.i.d., and hence the answer is $frac1n V X$.

Note that it depends on the i.i.d assumption, but not on the Poisson distribiton.

share|cite|improve this answer

answered Mar 29 at 3:41

Ott ToometOtt Toomet

1114

$endgroup$

add a comment | 

$begingroup$

I assume $tilde X$ is the same as $bar X$...

a) You are almost there but $E sum_i x_i = n E X$ given $X$ is i.i.d, hence the first answer $E bar X = E X$.

b) ditto here: $V sum_i X_i = n V X$ if $X$ is i.i.d., and hence the answer is $frac1n V X$.

Note that it depends on the i.i.d assumption, but not on the Poisson distribiton.

share|cite|improve this answer

answered Mar 29 at 3:41

Ott ToometOtt Toomet

1114

$endgroup$

share|cite|improve this answer

answered Mar 29 at 3:41

Ott ToometOtt Toomet

1114

answered Mar 29 at 3:41

Ott ToometOtt Toomet

1114

answered Mar 29 at 3:41

Ott ToometOtt Toomet

1114