Characteristic function of independent Poisson random variables
$begingroup$
Let $X_{i}$ be independent Poisson distributed random variables with parameter $lambda_{i} > 0$ for $i = 1,ldots,n$. Now the joint distribution is given by
begin{equation*}
mathbb{P}left(X_{1} = k_1,ldots,X_n = k_nright) = prod_{j=1}^{n}e^{-lambda_j}frac{lambda_j^{k_j}}{k_j!}.
end{equation*}
The characteristic function $varphi_{X_{i}}(t_{i})$ of $X_{i}$ is given by
$$
varphi_{X_{i}}(t_{i}) = e^{lambda_{i}left(e^{it_{i}} - 1 right)}.
$$
Is it right that the characteristic funtion of $(X_{1},ldots,X_n)$ is given by
$$
varphi_{X_{1}}(t_{1})cdotldotsvarphi_{X_{n}}(t_{n}) = e^{sum_{i=1}^{n}lambda_{i}left(e^{it_{i}} -1 right)}?
$$
probability poisson-distribution characteristic-functions
edited Jan 20 at 21:27
Davide Giraudo
128k17156268
asked Jan 16 at 19:29
love_mathlove_math
1615
$endgroup$
add a comment |
$begingroup$
Let $X_{i}$ be independent Poisson distributed random variables with parameter $lambda_{i} > 0$ for $i = 1,ldots,n$. Now the joint distribution is given by
begin{equation*}
mathbb{P}left(X_{1} = k_1,ldots,X_n = k_nright) = prod_{j=1}^{n}e^{-lambda_j}frac{lambda_j^{k_j}}{k_j!}.
end{equation*}
The characteristic function $varphi_{X_{i}}(t_{i})$ of $X_{i}$ is given by
$$
varphi_{X_{i}}(t_{i}) = e^{lambda_{i}left(e^{it_{i}} - 1 right)}.
$$
Is it right that the characteristic funtion of $(X_{1},ldots,X_n)$ is given by
$$
varphi_{X_{1}}(t_{1})cdotldotsvarphi_{X_{n}}(t_{n}) = e^{sum_{i=1}^{n}lambda_{i}left(e^{it_{i}} -1 right)}?
$$
probability poisson-distribution characteristic-functions
edited Jan 20 at 21:27
Davide Giraudo
128k17156268
asked Jan 16 at 19:29
love_mathlove_math
1615
$endgroup$
add a comment |
$begingroup$
Let $X_{i}$ be independent Poisson distributed random variables with parameter $lambda_{i} > 0$ for $i = 1,ldots,n$. Now the joint distribution is given by
begin{equation*}
mathbb{P}left(X_{1} = k_1,ldots,X_n = k_nright) = prod_{j=1}^{n}e^{-lambda_j}frac{lambda_j^{k_j}}{k_j!}.
end{equation*}
The characteristic function $varphi_{X_{i}}(t_{i})$ of $X_{i}$ is given by
$$
varphi_{X_{i}}(t_{i}) = e^{lambda_{i}left(e^{it_{i}} - 1 right)}.
$$
Is it right that the characteristic funtion of $(X_{1},ldots,X_n)$ is given by
$$
varphi_{X_{1}}(t_{1})cdotldotsvarphi_{X_{n}}(t_{n}) = e^{sum_{i=1}^{n}lambda_{i}left(e^{it_{i}} -1 right)}?
$$
probability poisson-distribution characteristic-functions
edited Jan 20 at 21:27
Davide Giraudo
128k17156268
asked Jan 16 at 19:29
love_mathlove_math
1615
$endgroup$
Let $X_{i}$ be independent Poisson distributed random variables with parameter $lambda_{i} > 0$ for $i = 1,ldots,n$. Now the joint distribution is given by
begin{equation*}
mathbb{P}left(X_{1} = k_1,ldots,X_n = k_nright) = prod_{j=1}^{n}e^{-lambda_j}frac{lambda_j^{k_j}}{k_j!}.
end{equation*}
The characteristic function $varphi_{X_{i}}(t_{i})$ of $X_{i}$ is given by
$$
varphi_{X_{i}}(t_{i}) = e^{lambda_{i}left(e^{it_{i}} - 1 right)}.
$$
Is it right that the characteristic funtion of $(X_{1},ldots,X_n)$ is given by
$$
varphi_{X_{1}}(t_{1})cdotldotsvarphi_{X_{n}}(t_{n}) = e^{sum_{i=1}^{n}lambda_{i}left(e^{it_{i}} -1 right)}?
$$
probability poisson-distribution characteristic-functions
probability poisson-distribution characteristic-functions
edited Jan 20 at 21:27
Davide Giraudo
128k17156268
asked Jan 16 at 19:29
love_mathlove_math
1615
edited Jan 20 at 21:27
Davide Giraudo
128k17156268
asked Jan 16 at 19:29
love_mathlove_math
1615
edited Jan 20 at 21:27
Davide Giraudo
128k17156268
edited Jan 20 at 21:27
Davide Giraudo
128k17156268
edited Jan 20 at 21:27
Davide Giraudo
128k17156268
128k17156268
asked Jan 16 at 19:29
love_mathlove_math
1615
asked Jan 16 at 19:29
love_mathlove_math
1615
asked Jan 16 at 19:29
love_mathlove_math
1615
1615
add a comment |
add a comment |
1 Answer
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Yes. The joint characteristic function of the $X_j$ is $Bbb Eexp itcdot X=Bbb Eprod_jexp it_jX_j$. By independence, $Bbb E$ commutes with $prod_j$.
answered Jan 16 at 19:50
J.G.J.G.
33.2k23252
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Yes. The joint characteristic function of the $X_j$ is $Bbb Eexp itcdot X=Bbb Eprod_jexp it_jX_j$. By independence, $Bbb E$ commutes with $prod_j$.
answered Jan 16 at 19:50
J.G.J.G.
33.2k23252
$endgroup$
add a comment |
$begingroup$
Yes. The joint characteristic function of the $X_j$ is $Bbb Eexp itcdot X=Bbb Eprod_jexp it_jX_j$. By independence, $Bbb E$ commutes with $prod_j$.
answered Jan 16 at 19:50
J.G.J.G.
33.2k23252
$endgroup$
add a comment |
$begingroup$
Yes. The joint characteristic function of the $X_j$ is $Bbb Eexp itcdot X=Bbb Eprod_jexp it_jX_j$. By independence, $Bbb E$ commutes with $prod_j$.
answered Jan 16 at 19:50
J.G.J.G.
33.2k23252
$endgroup$
Yes. The joint characteristic function of the $X_j$ is $Bbb Eexp itcdot X=Bbb Eprod_jexp it_jX_j$. By independence, $Bbb E$ commutes with $prod_j$.
answered Jan 16 at 19:50
J.G.J.G.
33.2k23252
answered Jan 16 at 19:50
J.G.J.G.
33.2k23252
answered Jan 16 at 19:50
J.G.J.G.
33.2k23252
answered Jan 16 at 19:50
J.G.J.G.
33.2k23252
33.2k23252
add a comment |
add a comment |
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