Limit of a mixture of binomial distributions
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Let $0 < r leq 1/2$ and let $B(n,p)$ be the notation for binomial distribution.
Consider the random variable $Z_n$ defined by the recursive equation
$$
Z_n+1 = Z_n + X_n I(Z_n geq r n^2) + Y_n I(Z_n < r n^2)
$$
and
$$
Z_1 sim B(1,r)
$$
where $X_m sim B(m,p_1)$ and $Y_m sim B(m,p_2)$ and $I(cdot)$ is the indicator function.
How to compute the probability distribution of $lim_n rightarrow infty fracZ_nn^2$
Thanks in advance.
markov-chains probability-limit-theorems large-deviation-theory
asked Aug 7 at 18:24
jaogye
425413
add a comment |Â
up vote
4
down vote
favorite
Let $0 < r leq 1/2$ and let $B(n,p)$ be the notation for binomial distribution.
Consider the random variable $Z_n$ defined by the recursive equation
$$
Z_n+1 = Z_n + X_n I(Z_n geq r n^2) + Y_n I(Z_n < r n^2)
$$
and
$$
Z_1 sim B(1,r)
$$
where $X_m sim B(m,p_1)$ and $Y_m sim B(m,p_2)$ and $I(cdot)$ is the indicator function.
How to compute the probability distribution of $lim_n rightarrow infty fracZ_nn^2$
Thanks in advance.
markov-chains probability-limit-theorems large-deviation-theory
asked Aug 7 at 18:24
jaogye
425413
Perhaps there is a typo, or I made a simple error: Since $E[Z_n] = O(n)$, it appears that $E[Q]=0$, where $Q$ is the RV whose distribution you wish to calculate. Since the pmf $f_q(q)=0$ for $q<0$, then does this not mean that pmf is $f_q(0)=1$ and zero otherwise?
– Dean
Aug 13 at 18:22
Dear @Dean, you are right $EZ_n] = O(n)$, so the limit $lim_n rightarrow infty fracZ_nn^2=0$. But what happend if $Z_n+1$ is the accumulation of previous $Z_n$, i.e. $Z_n+1 = Z_n + X_n I(Z_n geq r n^2) + Y_n I(Z_n < r n^2)$. I am going to change the question for a second time.
– jaogye
Aug 15 at 18:24
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Let $0 < r leq 1/2$ and let $B(n,p)$ be the notation for binomial distribution.
Consider the random variable $Z_n$ defined by the recursive equation
$$
Z_n+1 = Z_n + X_n I(Z_n geq r n^2) + Y_n I(Z_n < r n^2)
$$
and
$$
Z_1 sim B(1,r)
$$
where $X_m sim B(m,p_1)$ and $Y_m sim B(m,p_2)$ and $I(cdot)$ is the indicator function.
How to compute the probability distribution of $lim_n rightarrow infty fracZ_nn^2$
Thanks in advance.
markov-chains probability-limit-theorems large-deviation-theory
asked Aug 7 at 18:24
jaogye
425413
Let $0 < r leq 1/2$ and let $B(n,p)$ be the notation for binomial distribution.
Consider the random variable $Z_n$ defined by the recursive equation
$$
Z_n+1 = Z_n + X_n I(Z_n geq r n^2) + Y_n I(Z_n < r n^2)
$$
and
$$
Z_1 sim B(1,r)
$$
where $X_m sim B(m,p_1)$ and $Y_m sim B(m,p_2)$ and $I(cdot)$ is the indicator function.
How to compute the probability distribution of $lim_n rightarrow infty fracZ_nn^2$
Thanks in advance.
markov-chains probability-limit-theorems large-deviation-theory
asked Aug 7 at 18:24
jaogye
425413
edited Aug 15 at 18:27
asked Aug 7 at 18:24
jaogye
425413
asked Aug 7 at 18:24
jaogye
425413
asked Aug 7 at 18:24
jaogye
425413
425413
Perhaps there is a typo, or I made a simple error: Since $E[Z_n] = O(n)$, it appears that $E[Q]=0$, where $Q$ is the RV whose distribution you wish to calculate. Since the pmf $f_q(q)=0$ for $q<0$, then does this not mean that pmf is $f_q(0)=1$ and zero otherwise?
– Dean
Aug 13 at 18:22
Dear @Dean, you are right $EZ_n] = O(n)$, so the limit $lim_n rightarrow infty fracZ_nn^2=0$. But what happend if $Z_n+1$ is the accumulation of previous $Z_n$, i.e. $Z_n+1 = Z_n + X_n I(Z_n geq r n^2) + Y_n I(Z_n < r n^2)$. I am going to change the question for a second time.
– jaogye
Aug 15 at 18:24
add a comment |Â
Perhaps there is a typo, or I made a simple error: Since $E[Z_n] = O(n)$, it appears that $E[Q]=0$, where $Q$ is the RV whose distribution you wish to calculate. Since the pmf $f_q(q)=0$ for $q<0$, then does this not mean that pmf is $f_q(0)=1$ and zero otherwise?
– Dean
Aug 13 at 18:22
Dear @Dean, you are right $EZ_n] = O(n)$, so the limit $lim_n rightarrow infty fracZ_nn^2=0$. But what happend if $Z_n+1$ is the accumulation of previous $Z_n$, i.e. $Z_n+1 = Z_n + X_n I(Z_n geq r n^2) + Y_n I(Z_n < r n^2)$. I am going to change the question for a second time.
– jaogye
Aug 15 at 18:24
Perhaps there is a typo, or I made a simple error: Since $E[Z_n] = O(n)$, it appears that $E[Q]=0$, where $Q$ is the RV whose distribution you wish to calculate. Since the pmf $f_q(q)=0$ for $q<0$, then does this not mean that pmf is $f_q(0)=1$ and zero otherwise?
– Dean
Aug 13 at 18:22
Perhaps there is a typo, or I made a simple error: Since $E[Z_n] = O(n)$, it appears that $E[Q]=0$, where $Q$ is the RV whose distribution you wish to calculate. Since the pmf $f_q(q)=0$ for $q<0$, then does this not mean that pmf is $f_q(0)=1$ and zero otherwise?
– Dean
Aug 13 at 18:22
Dear @Dean, you are right $EZ_n] = O(n)$, so the limit $lim_n rightarrow infty fracZ_nn^2=0$. But what happend if $Z_n+1$ is the accumulation of previous $Z_n$, i.e. $Z_n+1 = Z_n + X_n I(Z_n geq r n^2) + Y_n I(Z_n < r n^2)$. I am going to change the question for a second time.
– jaogye
Aug 15 at 18:24
Dear @Dean, you are right $EZ_n] = O(n)$, so the limit $lim_n rightarrow infty fracZ_nn^2=0$. But what happend if $Z_n+1$ is the accumulation of previous $Z_n$, i.e. $Z_n+1 = Z_n + X_n I(Z_n geq r n^2) + Y_n I(Z_n < r n^2)$. I am going to change the question for a second time.
– jaogye
Aug 15 at 18:24
add a comment |Â
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Perhaps there is a typo, or I made a simple error: Since $E[Z_n] = O(n)$, it appears that $E[Q]=0$, where $Q$ is the RV whose distribution you wish to calculate. Since the pmf $f_q(q)=0$ for $q<0$, then does this not mean that pmf is $f_q(0)=1$ and zero otherwise?
– Dean
Aug 13 at 18:22
Dear @Dean, you are right $EZ_n] = O(n)$, so the limit $lim_n rightarrow infty fracZ_nn^2=0$. But what happend if $Z_n+1$ is the accumulation of previous $Z_n$, i.e. $Z_n+1 = Z_n + X_n I(Z_n geq r n^2) + Y_n I(Z_n < r n^2)$. I am going to change the question for a second time.
– jaogye
Aug 15 at 18:24