Show a given function is a characteristic function
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Let $phi(t)$ be a characteristic equation of a discrete r.v. X. Show that the function $e^mu (phi(t) -1)$ is a characteristic function. (Hint: Consider the sum of iid R.v.)
Attempt: I was told that the function could be shown to be a the sum of characteristic function of iid Poisson rv. I know the Poisson characteristic for the sum of 2 rv to be $e^(mu + lambda)(e^it-1)$. I am having trouble splitting this function $e^mu (phi(t) -1)$ to $e^(mu + lambda)(e^it-1)$. I was trying $e^mu (phi(t) -1) = sum frac(mu (phi (t)-1))^xx!$. But I got stuck as that's an infinite sum.
probability-theory
asked Aug 9 at 23:32
Dom
1587
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up vote
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Let $phi(t)$ be a characteristic equation of a discrete r.v. X. Show that the function $e^mu (phi(t) -1)$ is a characteristic function. (Hint: Consider the sum of iid R.v.)
Attempt: I was told that the function could be shown to be a the sum of characteristic function of iid Poisson rv. I know the Poisson characteristic for the sum of 2 rv to be $e^(mu + lambda)(e^it-1)$. I am having trouble splitting this function $e^mu (phi(t) -1)$ to $e^(mu + lambda)(e^it-1)$. I was trying $e^mu (phi(t) -1) = sum frac(mu (phi (t)-1))^xx!$. But I got stuck as that's an infinite sum.
probability-theory
asked Aug 9 at 23:32
Dom
1587
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $phi(t)$ be a characteristic equation of a discrete r.v. X. Show that the function $e^mu (phi(t) -1)$ is a characteristic function. (Hint: Consider the sum of iid R.v.)
Attempt: I was told that the function could be shown to be a the sum of characteristic function of iid Poisson rv. I know the Poisson characteristic for the sum of 2 rv to be $e^(mu + lambda)(e^it-1)$. I am having trouble splitting this function $e^mu (phi(t) -1)$ to $e^(mu + lambda)(e^it-1)$. I was trying $e^mu (phi(t) -1) = sum frac(mu (phi (t)-1))^xx!$. But I got stuck as that's an infinite sum.
probability-theory
asked Aug 9 at 23:32
Dom
1587
Let $phi(t)$ be a characteristic equation of a discrete r.v. X. Show that the function $e^mu (phi(t) -1)$ is a characteristic function. (Hint: Consider the sum of iid R.v.)
Attempt: I was told that the function could be shown to be a the sum of characteristic function of iid Poisson rv. I know the Poisson characteristic for the sum of 2 rv to be $e^(mu + lambda)(e^it-1)$. I am having trouble splitting this function $e^mu (phi(t) -1)$ to $e^(mu + lambda)(e^it-1)$. I was trying $e^mu (phi(t) -1) = sum frac(mu (phi (t)-1))^xx!$. But I got stuck as that's an infinite sum.
probability-theory
asked Aug 9 at 23:32
Dom
1587
asked Aug 9 at 23:32
Dom
1587
asked Aug 9 at 23:32
Dom
1587
asked Aug 9 at 23:32
Dom
1587
1587
add a comment |Â
add a comment |Â
1 Answer
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Let $X_i$ be i.i.d with characteristic function $phi$. Let $N$ be a Poisson random variable with parameter $mu$ independent of $X_i$'s. Then $Ee^it(X_1+X_2+...+X_N)=sum_n=0^infty Ee^it(X_1+X_2+...+X_n)e^-mu frac mu ^n n!=sum_n=0^infty phi (t)^ne^-mu frac mu ^n n!=e^mu (phi(t)-1)$.
answered Aug 9 at 23:38
Kavi Rama Murthy
21.6k2830
What is E here? The characteristic function? Shouldn't that be Phi?
– Dom
Aug 9 at 23:47
Sorry for typos. I have edited the answer.
– Kavi Rama Murthy
Aug 10 at 1:56
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Let $X_i$ be i.i.d with characteristic function $phi$. Let $N$ be a Poisson random variable with parameter $mu$ independent of $X_i$'s. Then $Ee^it(X_1+X_2+...+X_N)=sum_n=0^infty Ee^it(X_1+X_2+...+X_n)e^-mu frac mu ^n n!=sum_n=0^infty phi (t)^ne^-mu frac mu ^n n!=e^mu (phi(t)-1)$.
answered Aug 9 at 23:38
Kavi Rama Murthy
21.6k2830
What is E here? The characteristic function? Shouldn't that be Phi?
– Dom
Aug 9 at 23:47
Sorry for typos. I have edited the answer.
– Kavi Rama Murthy
Aug 10 at 1:56
add a comment |Â
up vote
1
down vote
accepted
Let $X_i$ be i.i.d with characteristic function $phi$. Let $N$ be a Poisson random variable with parameter $mu$ independent of $X_i$'s. Then $Ee^it(X_1+X_2+...+X_N)=sum_n=0^infty Ee^it(X_1+X_2+...+X_n)e^-mu frac mu ^n n!=sum_n=0^infty phi (t)^ne^-mu frac mu ^n n!=e^mu (phi(t)-1)$.
answered Aug 9 at 23:38
Kavi Rama Murthy
21.6k2830
What is E here? The characteristic function? Shouldn't that be Phi?
– Dom
Aug 9 at 23:47
Sorry for typos. I have edited the answer.
– Kavi Rama Murthy
Aug 10 at 1:56
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Let $X_i$ be i.i.d with characteristic function $phi$. Let $N$ be a Poisson random variable with parameter $mu$ independent of $X_i$'s. Then $Ee^it(X_1+X_2+...+X_N)=sum_n=0^infty Ee^it(X_1+X_2+...+X_n)e^-mu frac mu ^n n!=sum_n=0^infty phi (t)^ne^-mu frac mu ^n n!=e^mu (phi(t)-1)$.
answered Aug 9 at 23:38
Kavi Rama Murthy
21.6k2830
Let $X_i$ be i.i.d with characteristic function $phi$. Let $N$ be a Poisson random variable with parameter $mu$ independent of $X_i$'s. Then $Ee^it(X_1+X_2+...+X_N)=sum_n=0^infty Ee^it(X_1+X_2+...+X_n)e^-mu frac mu ^n n!=sum_n=0^infty phi (t)^ne^-mu frac mu ^n n!=e^mu (phi(t)-1)$.
answered Aug 9 at 23:38
Kavi Rama Murthy
21.6k2830
edited Aug 10 at 1:58
answered Aug 9 at 23:38
Kavi Rama Murthy
21.6k2830
answered Aug 9 at 23:38
Kavi Rama Murthy
21.6k2830
answered Aug 9 at 23:38
Kavi Rama Murthy
21.6k2830
21.6k2830
What is E here? The characteristic function? Shouldn't that be Phi?
– Dom
Aug 9 at 23:47
Sorry for typos. I have edited the answer.
– Kavi Rama Murthy
Aug 10 at 1:56
add a comment |Â
What is E here? The characteristic function? Shouldn't that be Phi?
– Dom
Aug 9 at 23:47
Sorry for typos. I have edited the answer.
– Kavi Rama Murthy
Aug 10 at 1:56
What is E here? The characteristic function? Shouldn't that be Phi?
– Dom
Aug 9 at 23:47
What is E here? The characteristic function? Shouldn't that be Phi?
– Dom
Aug 9 at 23:47
Sorry for typos. I have edited the answer.
– Kavi Rama Murthy
Aug 10 at 1:56
Sorry for typos. I have edited the answer.
– Kavi Rama Murthy
Aug 10 at 1:56
add a comment |Â
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