Suppose $frac1sqrtnsum_i=1^nY_ioversetdto N(0,V).$ What is the distribution of$frac1sqrtnsum_i=1^nG(Y_i)$? The Next CEO of Stack OverflowLimiting distribution $displaystylefracsum_i=1^n X_isum_i=1^n Y_i$Use $2sum_i=1^n Y_i/beta$ which is a pivotal quantity to derive a 95% confidence interval for $beta$Weak Law of large numbers involving a sequence and random variableDeriving the asymptotic distribution of a two-stage estimatorFor $Y_1, ldots, Y_n$ i.i.d. with Poisson distribution, and $S_n=sumlimits_i=1^nY_i$, how to show $E(Y_1 mid S_n) = frac1nS_n$?In the CLT, $fracsum_i=1^nX_is_noversetdto N(0,1)$ implies $fracX_ns_noversetdto N(0,rho^2).$Suppose that $sqrtn(X_n-X) oversetD to mathcalN(0, sigma^2)$. What does $(sqrtn(X_n-X))^2$ converge in distribution to?Difference between $frac sum_i=1^n (Y_i - barY)^2n$ and $frac sum_i=1^n (Y_i - barY)^2n-1$Distribution of $2k fracZ^2sum_i=1^k Y_i$, where $Z sim operatornameN(0,1)$ and $Y sim operatornameEXP(2)$Digits: Convergence of number of $00$
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Suppose $frac1sqrtnsum_i=1^nY_ioversetdto N(0,V).$ What is the distribution of$frac1sqrtnsum_i=1^nG(Y_i)$?
The Next CEO of Stack OverflowLimiting distribution $displaystylefracsum_i=1^n X_isum_i=1^n Y_i$Use $2sum_i=1^n Y_i/beta$ which is a pivotal quantity to derive a 95% confidence interval for $beta$Weak Law of large numbers involving a sequence and random variableDeriving the asymptotic distribution of a two-stage estimatorFor $Y_1, ldots, Y_n$ i.i.d. with Poisson distribution, and $S_n=sumlimits_i=1^nY_i$, how to show $E(Y_1 mid S_n) = frac1nS_n$?In the CLT, $fracsum_i=1^nX_is_noversetdto N(0,1)$ implies $fracX_ns_noversetdto N(0,rho^2).$Suppose that $sqrtn(X_n-X) oversetD to mathcalN(0, sigma^2)$. What does $(sqrtn(X_n-X))^2$ converge in distribution to?Difference between $frac sum_i=1^n (Y_i - barY)^2n$ and $frac sum_i=1^n (Y_i - barY)^2n-1$Distribution of $2k fracZ^2sum_i=1^k Y_i$, where $Z sim operatornameN(0,1)$ and $Y sim operatornameEXP(2)$Digits: Convergence of number of $00$
$begingroup$
Suppose
$$frac1sqrtnsum_i=1^nY_ioversetdto N(0,V).$$
Let $G(x)=int_-infty^xk(u)du$ be a kernel distribution function.
Can we obtain the asymptotic distribution of $frac1sqrtnsum_i=1^nG(Y_i)$?
Another question is if
$$frac1sqrtnsum_i=1^nY_ioversetdto N(0,V).$$
Can we obtain that
$frac1nsum_i=1^nY_i^2oversetpto V$?
probability-theory probability-distributions law-of-large-numbers
asked Mar 27 at 18:13
J.MikeJ.Mike
336110
$endgroup$
add a comment |
$begingroup$
Suppose
$$frac1sqrtnsum_i=1^nY_ioversetdto N(0,V).$$
Let $G(x)=int_-infty^xk(u)du$ be a kernel distribution function.
Can we obtain the asymptotic distribution of $frac1sqrtnsum_i=1^nG(Y_i)$?
Another question is if
$$frac1sqrtnsum_i=1^nY_ioversetdto N(0,V).$$
Can we obtain that
$frac1nsum_i=1^nY_i^2oversetpto V$?
probability-theory probability-distributions law-of-large-numbers
asked Mar 27 at 18:13
J.MikeJ.Mike
336110
$endgroup$
$begingroup$
For the first question Im not sure, but assuming that $G(0)=0$ and $G'(x)=k(x)neq 0$, then using the delta method you have $ 1/sqrtn sum G(Y_i) xrightarrowD N(0, k'(0)^2V)$
$endgroup$
– V. Vancak
2 days ago
$begingroup$
Thank you for your kind comment. I think it is slightly different from delta method. $frac1sqrtnsum_i=1^nY_i=sqrtncdotfrac1nsum_i=1^nY_i=sqrtn(barY-0)oversetdto N(0,V)$ Using delta method, we obtain $sqrtn(G(barY)-G(0))oversetdto N(0,[G^prime(0)]^2cdot V)$. Generally speaking, $G(frac1nsum_i=1^nY_i)neq frac1nsum_i=1^nG(Y_i)$.
$endgroup$
– J.Mike
2 days ago
$begingroup$
For iid case, if $mathsf EG(Y_1)neq 0$ then $frac1sqrtnsum_i=1^nG(Y_i)$ does not have limiting distribution. It diverges by LLN. Do you have any restrictions that make this expectation zero?
$endgroup$
– NCh
2 days ago
$begingroup$
I check the model, there is a moment conditon saying that $EG(Y_1)=0$. Sorry about that.
$endgroup$
– J.Mike
2 days ago
add a comment |
$begingroup$
Suppose
$$frac1sqrtnsum_i=1^nY_ioversetdto N(0,V).$$
Let $G(x)=int_-infty^xk(u)du$ be a kernel distribution function.
Can we obtain the asymptotic distribution of $frac1sqrtnsum_i=1^nG(Y_i)$?
Another question is if
$$frac1sqrtnsum_i=1^nY_ioversetdto N(0,V).$$
Can we obtain that
$frac1nsum_i=1^nY_i^2oversetpto V$?
probability-theory probability-distributions law-of-large-numbers
asked Mar 27 at 18:13
J.MikeJ.Mike
336110
$endgroup$
Suppose
$$frac1sqrtnsum_i=1^nY_ioversetdto N(0,V).$$
Let $G(x)=int_-infty^xk(u)du$ be a kernel distribution function.
Can we obtain the asymptotic distribution of $frac1sqrtnsum_i=1^nG(Y_i)$?
Another question is if
$$frac1sqrtnsum_i=1^nY_ioversetdto N(0,V).$$
Can we obtain that
$frac1nsum_i=1^nY_i^2oversetpto V$?
probability-theory probability-distributions law-of-large-numbers
probability-theory probability-distributions law-of-large-numbers
asked Mar 27 at 18:13
J.MikeJ.Mike
336110
asked Mar 27 at 18:13
J.MikeJ.Mike
336110
edited Mar 27 at 18:21
J.Mike
asked Mar 27 at 18:13
J.MikeJ.Mike
336110
asked Mar 27 at 18:13
J.MikeJ.Mike
336110
asked Mar 27 at 18:13
J.MikeJ.Mike
336110
336110
$begingroup$
For the first question Im not sure, but assuming that $G(0)=0$ and $G'(x)=k(x)neq 0$, then using the delta method you have $ 1/sqrtn sum G(Y_i) xrightarrowD N(0, k'(0)^2V)$
$endgroup$
– V. Vancak
2 days ago
$begingroup$
Thank you for your kind comment. I think it is slightly different from delta method. $frac1sqrtnsum_i=1^nY_i=sqrtncdotfrac1nsum_i=1^nY_i=sqrtn(barY-0)oversetdto N(0,V)$ Using delta method, we obtain $sqrtn(G(barY)-G(0))oversetdto N(0,[G^prime(0)]^2cdot V)$. Generally speaking, $G(frac1nsum_i=1^nY_i)neq frac1nsum_i=1^nG(Y_i)$.
$endgroup$
– J.Mike
2 days ago
$begingroup$
For iid case, if $mathsf EG(Y_1)neq 0$ then $frac1sqrtnsum_i=1^nG(Y_i)$ does not have limiting distribution. It diverges by LLN. Do you have any restrictions that make this expectation zero?
$endgroup$
– NCh
2 days ago
$begingroup$
I check the model, there is a moment conditon saying that $EG(Y_1)=0$. Sorry about that.
$endgroup$
– J.Mike
2 days ago
add a comment |
$begingroup$
For the first question Im not sure, but assuming that $G(0)=0$ and $G'(x)=k(x)neq 0$, then using the delta method you have $ 1/sqrtn sum G(Y_i) xrightarrowD N(0, k'(0)^2V)$
$endgroup$
– V. Vancak
2 days ago
$begingroup$
Thank you for your kind comment. I think it is slightly different from delta method. $frac1sqrtnsum_i=1^nY_i=sqrtncdotfrac1nsum_i=1^nY_i=sqrtn(barY-0)oversetdto N(0,V)$ Using delta method, we obtain $sqrtn(G(barY)-G(0))oversetdto N(0,[G^prime(0)]^2cdot V)$. Generally speaking, $G(frac1nsum_i=1^nY_i)neq frac1nsum_i=1^nG(Y_i)$.
$endgroup$
– J.Mike
2 days ago
$begingroup$
For iid case, if $mathsf EG(Y_1)neq 0$ then $frac1sqrtnsum_i=1^nG(Y_i)$ does not have limiting distribution. It diverges by LLN. Do you have any restrictions that make this expectation zero?
$endgroup$
– NCh
2 days ago
$begingroup$
I check the model, there is a moment conditon saying that $EG(Y_1)=0$. Sorry about that.
$endgroup$
– J.Mike
2 days ago
$begingroup$
For the first question Im not sure, but assuming that $G(0)=0$ and $G'(x)=k(x)neq 0$, then using the delta method you have $ 1/sqrtn sum G(Y_i) xrightarrowD N(0, k'(0)^2V)$
$endgroup$
– V. Vancak
2 days ago
$begingroup$
For the first question Im not sure, but assuming that $G(0)=0$ and $G'(x)=k(x)neq 0$, then using the delta method you have $ 1/sqrtn sum G(Y_i) xrightarrowD N(0, k'(0)^2V)$
$endgroup$
– V. Vancak
2 days ago
$begingroup$
Thank you for your kind comment. I think it is slightly different from delta method. $frac1sqrtnsum_i=1^nY_i=sqrtncdotfrac1nsum_i=1^nY_i=sqrtn(barY-0)oversetdto N(0,V)$ Using delta method, we obtain $sqrtn(G(barY)-G(0))oversetdto N(0,[G^prime(0)]^2cdot V)$. Generally speaking, $G(frac1nsum_i=1^nY_i)neq frac1nsum_i=1^nG(Y_i)$.
$endgroup$
– J.Mike
2 days ago
$begingroup$
Thank you for your kind comment. I think it is slightly different from delta method. $frac1sqrtnsum_i=1^nY_i=sqrtncdotfrac1nsum_i=1^nY_i=sqrtn(barY-0)oversetdto N(0,V)$ Using delta method, we obtain $sqrtn(G(barY)-G(0))oversetdto N(0,[G^prime(0)]^2cdot V)$. Generally speaking, $G(frac1nsum_i=1^nY_i)neq frac1nsum_i=1^nG(Y_i)$.
$endgroup$
– J.Mike
2 days ago
$begingroup$
For iid case, if $mathsf EG(Y_1)neq 0$ then $frac1sqrtnsum_i=1^nG(Y_i)$ does not have limiting distribution. It diverges by LLN. Do you have any restrictions that make this expectation zero?
$endgroup$
– NCh
2 days ago
$begingroup$
For iid case, if $mathsf EG(Y_1)neq 0$ then $frac1sqrtnsum_i=1^nG(Y_i)$ does not have limiting distribution. It diverges by LLN. Do you have any restrictions that make this expectation zero?
$endgroup$
– NCh
2 days ago
$begingroup$
I check the model, there is a moment conditon saying that $EG(Y_1)=0$. Sorry about that.
$endgroup$
– J.Mike
2 days ago
$begingroup$
I check the model, there is a moment conditon saying that $EG(Y_1)=0$. Sorry about that.
$endgroup$
– J.Mike
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For the second question - assuming i.i.d and finite fourth moment, using the WLLN you have
$$
frac1nsum_i Y_i^2xrightarrowpmathbbEY^2,
$$
where
$$
mathbbEY^2=Var(Y)+mathbbE^2Y=V+0=V.
$$
answered 2 days ago
V. VancakV. Vancak
11.4k3926
$endgroup$
add a comment |
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1 Answer
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oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For the second question - assuming i.i.d and finite fourth moment, using the WLLN you have
$$
frac1nsum_i Y_i^2xrightarrowpmathbbEY^2,
$$
where
$$
mathbbEY^2=Var(Y)+mathbbE^2Y=V+0=V.
$$
answered 2 days ago
V. VancakV. Vancak
11.4k3926
$endgroup$
add a comment |
$begingroup$
For the second question - assuming i.i.d and finite fourth moment, using the WLLN you have
$$
frac1nsum_i Y_i^2xrightarrowpmathbbEY^2,
$$
where
$$
mathbbEY^2=Var(Y)+mathbbE^2Y=V+0=V.
$$
answered 2 days ago
V. VancakV. Vancak
11.4k3926
$endgroup$
add a comment |
$begingroup$
For the second question - assuming i.i.d and finite fourth moment, using the WLLN you have
$$
frac1nsum_i Y_i^2xrightarrowpmathbbEY^2,
$$
where
$$
mathbbEY^2=Var(Y)+mathbbE^2Y=V+0=V.
$$
answered 2 days ago
V. VancakV. Vancak
11.4k3926
$endgroup$
For the second question - assuming i.i.d and finite fourth moment, using the WLLN you have
$$
frac1nsum_i Y_i^2xrightarrowpmathbbEY^2,
$$
where
$$
mathbbEY^2=Var(Y)+mathbbE^2Y=V+0=V.
$$
answered 2 days ago
V. VancakV. Vancak
11.4k3926
answered 2 days ago
V. VancakV. Vancak
11.4k3926
answered 2 days ago
V. VancakV. Vancak
11.4k3926
answered 2 days ago
V. VancakV. Vancak
11.4k3926
11.4k3926
add a comment |
add a comment |
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$begingroup$
For the first question Im not sure, but assuming that $G(0)=0$ and $G'(x)=k(x)neq 0$, then using the delta method you have $ 1/sqrtn sum G(Y_i) xrightarrowD N(0, k'(0)^2V)$
$endgroup$
– V. Vancak
2 days ago
$begingroup$
Thank you for your kind comment. I think it is slightly different from delta method. $frac1sqrtnsum_i=1^nY_i=sqrtncdotfrac1nsum_i=1^nY_i=sqrtn(barY-0)oversetdto N(0,V)$ Using delta method, we obtain $sqrtn(G(barY)-G(0))oversetdto N(0,[G^prime(0)]^2cdot V)$. Generally speaking, $G(frac1nsum_i=1^nY_i)neq frac1nsum_i=1^nG(Y_i)$.
$endgroup$
– J.Mike
2 days ago
$begingroup$
For iid case, if $mathsf EG(Y_1)neq 0$ then $frac1sqrtnsum_i=1^nG(Y_i)$ does not have limiting distribution. It diverges by LLN. Do you have any restrictions that make this expectation zero?
$endgroup$
– NCh
2 days ago
$begingroup$
I check the model, there is a moment conditon saying that $EG(Y_1)=0$. Sorry about that.
$endgroup$
– J.Mike
2 days ago