Mixing
$X_n$ ~ Poisson(n), $Y_n$ ~ Geometric($e^{-frac{1}{n}})$
$begingroup$
Let $X_n$ ~ Poisson(n), $Y_n$ ~ Geometric($e^{-frac{1}{n}})$, everything independent. I want to find the convergence in law of:
$$ Z_n = frac{1}{n}X_n + beta Y_n $$
With $beta in mathbb{R}$.
Now, my plan is to use the characteristic function of $Z_n$.
This is what I have done:
$$phi_{Z_n} = phi_{frac{1}{n}X_n} *phi_{beta Y_n} $$
$$phi_{Z_n} = frac{ e^{-frac{1}{n}} e^{ibeta t}}{1-(1-e^{-frac{1}{n}})e^{ibeta t} } *e^{n(e^{it} - 1)}$$
I hope my calculations make sense up untili this point. My problem is now calculating the limit of n going to infinity. Any idea how to solve this?
probability weak-convergence
asked Jan 22 at 14:46
qcc101qcc101
627213
$endgroup$
add a comment |
$begingroup$
Let $X_n$ ~ Poisson(n), $Y_n$ ~ Geometric($e^{-frac{1}{n}})$, everything independent. I want to find the convergence in law of:
$$ Z_n = frac{1}{n}X_n + beta Y_n $$
With $beta in mathbb{R}$.
Now, my plan is to use the characteristic function of $Z_n$.
This is what I have done:
$$phi_{Z_n} = phi_{frac{1}{n}X_n} *phi_{beta Y_n} $$
$$phi_{Z_n} = frac{ e^{-frac{1}{n}} e^{ibeta t}}{1-(1-e^{-frac{1}{n}})e^{ibeta t} } *e^{n(e^{it} - 1)}$$
I hope my calculations make sense up untili this point. My problem is now calculating the limit of n going to infinity. Any idea how to solve this?
probability weak-convergence
asked Jan 22 at 14:46
qcc101qcc101
627213
$endgroup$
$begingroup$
Is your fraction just 1? Is there an exponent missing?
$endgroup$
– Paul
Jan 22 at 14:50
$begingroup$
@Paul Edited, thank you for catching the mistake.
$endgroup$
– qcc101
Jan 22 at 15:06
add a comment |
$begingroup$
Let $X_n$ ~ Poisson(n), $Y_n$ ~ Geometric($e^{-frac{1}{n}})$, everything independent. I want to find the convergence in law of:
$$ Z_n = frac{1}{n}X_n + beta Y_n $$
With $beta in mathbb{R}$.
Now, my plan is to use the characteristic function of $Z_n$.
This is what I have done:
$$phi_{Z_n} = phi_{frac{1}{n}X_n} *phi_{beta Y_n} $$
$$phi_{Z_n} = frac{ e^{-frac{1}{n}} e^{ibeta t}}{1-(1-e^{-frac{1}{n}})e^{ibeta t} } *e^{n(e^{it} - 1)}$$
I hope my calculations make sense up untili this point. My problem is now calculating the limit of n going to infinity. Any idea how to solve this?
probability weak-convergence
asked Jan 22 at 14:46
qcc101qcc101
627213
$endgroup$
Let $X_n$ ~ Poisson(n), $Y_n$ ~ Geometric($e^{-frac{1}{n}})$, everything independent. I want to find the convergence in law of:
$$ Z_n = frac{1}{n}X_n + beta Y_n $$
With $beta in mathbb{R}$.
Now, my plan is to use the characteristic function of $Z_n$.
This is what I have done:
$$phi_{Z_n} = phi_{frac{1}{n}X_n} *phi_{beta Y_n} $$
$$phi_{Z_n} = frac{ e^{-frac{1}{n}} e^{ibeta t}}{1-(1-e^{-frac{1}{n}})e^{ibeta t} } *e^{n(e^{it} - 1)}$$
I hope my calculations make sense up untili this point. My problem is now calculating the limit of n going to infinity. Any idea how to solve this?
probability weak-convergence
probability weak-convergence
asked Jan 22 at 14:46
qcc101qcc101
627213
asked Jan 22 at 14:46
qcc101qcc101
627213
edited Jan 22 at 15:06
qcc101
asked Jan 22 at 14:46
qcc101qcc101
627213
asked Jan 22 at 14:46
qcc101qcc101
627213
asked Jan 22 at 14:46
qcc101qcc101
627213
627213
$begingroup$
Is your fraction just 1? Is there an exponent missing?
$endgroup$
– Paul
Jan 22 at 14:50
$begingroup$
@Paul Edited, thank you for catching the mistake.
$endgroup$
– qcc101
Jan 22 at 15:06
add a comment |
$begingroup$
Is your fraction just 1? Is there an exponent missing?
$endgroup$
– Paul
Jan 22 at 14:50
$begingroup$
@Paul Edited, thank you for catching the mistake.
$endgroup$
– qcc101
Jan 22 at 15:06
$begingroup$
Is your fraction just 1? Is there an exponent missing?
$endgroup$
– Paul
Jan 22 at 14:50
$begingroup$
Is your fraction just 1? Is there an exponent missing?
$endgroup$
– Paul
Jan 22 at 14:50
$begingroup$
@Paul Edited, thank you for catching the mistake.
$endgroup$
– qcc101
Jan 22 at 15:06
$begingroup$
@Paul Edited, thank you for catching the mistake.
$endgroup$
– qcc101
Jan 22 at 15:06
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Note that defining $(W_n)_{n in mathbb{N}}$ where each $W_n$ is independent and Poisson distributed with mean $1$, we get
$$
frac{1}{n} X_n overset{mathcal{D}}{=} frac{1}{n} sum_{i=1}^n W_i
$$
and it is obvious from the law of large numbers that this converges to $1$ almost surely (hence in probability and in distribution).
Chebyshev's inequality can be used to prove that $beta Y_n to beta$ in probability since for any $varepsilon > 0$
$$
P(|beta Y_n - beta| geq varepsilon) = Pleft(|Y-1| geq frac{varepsilon}{|beta|} right) leq frac{textrm{Var}(Y_n) beta^2}{varepsilon^2}
$$
and since $textrm{Var}(Y_n) = frac{1-e^{-frac{1}{n}}}{(e^{-frac{1}{n}})^2 } to 0$ as $n to infty$, we are done. (This argument could also be used for the Poisson variables)
Convergence in probability is stable under summation and implies convergence in distribution.
answered Jan 22 at 15:33
LundborgLundborg
852517
$endgroup$
$begingroup$
Thank you, super clear. What if I wanted to use the characteristic function though?
$endgroup$
– qcc101
Jan 22 at 15:42
$begingroup$
@qcc101 I don't really see why you would want that when this is easier and proves something stronger but note that you're missing a $1/n$ in your expression for the characteristic function for $1/n X_n$. It should be $exp(n(e^{frac{it}{n}}-1))$. I think this helps showing the same result as above by applying the usual laws for limits.
$endgroup$
– Lundborg
Jan 22 at 16:25
add a comment |
Your Answer
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
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active
oldest
votes
active
oldest
votes
$begingroup$
Note that defining $(W_n)_{n in mathbb{N}}$ where each $W_n$ is independent and Poisson distributed with mean $1$, we get
$$
frac{1}{n} X_n overset{mathcal{D}}{=} frac{1}{n} sum_{i=1}^n W_i
$$
and it is obvious from the law of large numbers that this converges to $1$ almost surely (hence in probability and in distribution).
Chebyshev's inequality can be used to prove that $beta Y_n to beta$ in probability since for any $varepsilon > 0$
$$
P(|beta Y_n - beta| geq varepsilon) = Pleft(|Y-1| geq frac{varepsilon}{|beta|} right) leq frac{textrm{Var}(Y_n) beta^2}{varepsilon^2}
$$
and since $textrm{Var}(Y_n) = frac{1-e^{-frac{1}{n}}}{(e^{-frac{1}{n}})^2 } to 0$ as $n to infty$, we are done. (This argument could also be used for the Poisson variables)
Convergence in probability is stable under summation and implies convergence in distribution.
answered Jan 22 at 15:33
LundborgLundborg
852517
$endgroup$
$begingroup$
Thank you, super clear. What if I wanted to use the characteristic function though?
$endgroup$
– qcc101
Jan 22 at 15:42
$begingroup$
@qcc101 I don't really see why you would want that when this is easier and proves something stronger but note that you're missing a $1/n$ in your expression for the characteristic function for $1/n X_n$. It should be $exp(n(e^{frac{it}{n}}-1))$. I think this helps showing the same result as above by applying the usual laws for limits.
$endgroup$
– Lundborg
Jan 22 at 16:25
add a comment |
$begingroup$
Note that defining $(W_n)_{n in mathbb{N}}$ where each $W_n$ is independent and Poisson distributed with mean $1$, we get
$$
frac{1}{n} X_n overset{mathcal{D}}{=} frac{1}{n} sum_{i=1}^n W_i
$$
and it is obvious from the law of large numbers that this converges to $1$ almost surely (hence in probability and in distribution).
Chebyshev's inequality can be used to prove that $beta Y_n to beta$ in probability since for any $varepsilon > 0$
$$
P(|beta Y_n - beta| geq varepsilon) = Pleft(|Y-1| geq frac{varepsilon}{|beta|} right) leq frac{textrm{Var}(Y_n) beta^2}{varepsilon^2}
$$
and since $textrm{Var}(Y_n) = frac{1-e^{-frac{1}{n}}}{(e^{-frac{1}{n}})^2 } to 0$ as $n to infty$, we are done. (This argument could also be used for the Poisson variables)
Convergence in probability is stable under summation and implies convergence in distribution.
answered Jan 22 at 15:33
LundborgLundborg
852517
$endgroup$
$begingroup$
Thank you, super clear. What if I wanted to use the characteristic function though?
$endgroup$
– qcc101
Jan 22 at 15:42
$begingroup$
@qcc101 I don't really see why you would want that when this is easier and proves something stronger but note that you're missing a $1/n$ in your expression for the characteristic function for $1/n X_n$. It should be $exp(n(e^{frac{it}{n}}-1))$. I think this helps showing the same result as above by applying the usual laws for limits.
$endgroup$
– Lundborg
Jan 22 at 16:25
add a comment |
$begingroup$
Note that defining $(W_n)_{n in mathbb{N}}$ where each $W_n$ is independent and Poisson distributed with mean $1$, we get
$$
frac{1}{n} X_n overset{mathcal{D}}{=} frac{1}{n} sum_{i=1}^n W_i
$$
and it is obvious from the law of large numbers that this converges to $1$ almost surely (hence in probability and in distribution).
Chebyshev's inequality can be used to prove that $beta Y_n to beta$ in probability since for any $varepsilon > 0$
$$
P(|beta Y_n - beta| geq varepsilon) = Pleft(|Y-1| geq frac{varepsilon}{|beta|} right) leq frac{textrm{Var}(Y_n) beta^2}{varepsilon^2}
$$
and since $textrm{Var}(Y_n) = frac{1-e^{-frac{1}{n}}}{(e^{-frac{1}{n}})^2 } to 0$ as $n to infty$, we are done. (This argument could also be used for the Poisson variables)
Convergence in probability is stable under summation and implies convergence in distribution.
answered Jan 22 at 15:33
LundborgLundborg
852517
$endgroup$
Note that defining $(W_n)_{n in mathbb{N}}$ where each $W_n$ is independent and Poisson distributed with mean $1$, we get
$$
frac{1}{n} X_n overset{mathcal{D}}{=} frac{1}{n} sum_{i=1}^n W_i
$$
and it is obvious from the law of large numbers that this converges to $1$ almost surely (hence in probability and in distribution).
Chebyshev's inequality can be used to prove that $beta Y_n to beta$ in probability since for any $varepsilon > 0$
$$
P(|beta Y_n - beta| geq varepsilon) = Pleft(|Y-1| geq frac{varepsilon}{|beta|} right) leq frac{textrm{Var}(Y_n) beta^2}{varepsilon^2}
$$
and since $textrm{Var}(Y_n) = frac{1-e^{-frac{1}{n}}}{(e^{-frac{1}{n}})^2 } to 0$ as $n to infty$, we are done. (This argument could also be used for the Poisson variables)
Convergence in probability is stable under summation and implies convergence in distribution.
answered Jan 22 at 15:33
LundborgLundborg
852517
answered Jan 22 at 15:33
LundborgLundborg
852517
answered Jan 22 at 15:33
LundborgLundborg
852517
answered Jan 22 at 15:33
LundborgLundborg
852517
852517
$begingroup$
Thank you, super clear. What if I wanted to use the characteristic function though?
$endgroup$
– qcc101
Jan 22 at 15:42
$begingroup$
@qcc101 I don't really see why you would want that when this is easier and proves something stronger but note that you're missing a $1/n$ in your expression for the characteristic function for $1/n X_n$. It should be $exp(n(e^{frac{it}{n}}-1))$. I think this helps showing the same result as above by applying the usual laws for limits.
$endgroup$
– Lundborg
Jan 22 at 16:25
add a comment |
$begingroup$
Thank you, super clear. What if I wanted to use the characteristic function though?
$endgroup$
– qcc101
Jan 22 at 15:42
$begingroup$
@qcc101 I don't really see why you would want that when this is easier and proves something stronger but note that you're missing a $1/n$ in your expression for the characteristic function for $1/n X_n$. It should be $exp(n(e^{frac{it}{n}}-1))$. I think this helps showing the same result as above by applying the usual laws for limits.
$endgroup$
– Lundborg
Jan 22 at 16:25
$begingroup$
Thank you, super clear. What if I wanted to use the characteristic function though?
$endgroup$
– qcc101
Jan 22 at 15:42
$begingroup$
Thank you, super clear. What if I wanted to use the characteristic function though?
$endgroup$
– qcc101
Jan 22 at 15:42
$begingroup$
@qcc101 I don't really see why you would want that when this is easier and proves something stronger but note that you're missing a $1/n$ in your expression for the characteristic function for $1/n X_n$. It should be $exp(n(e^{frac{it}{n}}-1))$. I think this helps showing the same result as above by applying the usual laws for limits.
$endgroup$
– Lundborg
Jan 22 at 16:25
$begingroup$
@qcc101 I don't really see why you would want that when this is easier and proves something stronger but note that you're missing a $1/n$ in your expression for the characteristic function for $1/n X_n$. It should be $exp(n(e^{frac{it}{n}}-1))$. I think this helps showing the same result as above by applying the usual laws for limits.
$endgroup$
– Lundborg
Jan 22 at 16:25
add a comment |
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$begingroup$
Is your fraction just 1? Is there an exponent missing?
$endgroup$
– Paul
Jan 22 at 14:50
$begingroup$
@Paul Edited, thank you for catching the mistake.
$endgroup$
– qcc101
Jan 22 at 15:06