$X_n$ ~ Poisson(n), $Y_n$ ~ Geometric($e^{-frac{1}{n}})$












1












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















1












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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














1












1








1





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 22 at 15:06







qcc101

















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


















  • $begingroup$
    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










1 Answer
1






active

oldest

votes


















1












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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













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1 Answer
1






active

oldest

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1 Answer
1






active

oldest

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active

oldest

votes






active

oldest

votes









1












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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


















1












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















1












1








1





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 22 at 15:33









LundborgLundborg

880416




880416












  • $begingroup$
    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










  • $begingroup$
    @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




















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