Mixing
Central limit theorem (the method of “moments”)
$begingroup$
I'm trying to complete this problem. It's been years that I studied probability, so if someone could point me in the right direction, I'd appreciate it.
Let $Z_1, Z_2, ...$ be a sequence of IID random variables with mean $0$ and variance $1$ and define
begin{align*}
S_n = frac{1}{sqrt{n}}sum_{i=1}^{n}Z_i
end{align*}
and
begin{align*}
L_m=lim_{n to infty}E[S_{n}^{m}].
end{align*}
Another method of proof of the Central Limit theorem works by showing that for each $m$, the limit $L_m$ exists, and the sequence satisfies the recurrence relation
begin{align*}
L_{m+1}=mL_{m-1}.
end{align*}
1) Use integration by parts to show that the sequence $R_m=E[Z^{m}]$, where $Z$ is a normal $N(0,1)$ random variable, satisfies the same recurrence relation.
2) Use induction to deduce that the above recurrence relation implies $L_m=R_m=0$ if $m$ is odd, and that they are both equal to
begin{align*}
frac{m!}{2^{m/2}(m/2)!}
end{align*}
if $m$ is even.
Any guidance at all would be appreciated. I'm not sure where to begin here.
probability-theory
asked Jan 12 at 14:22
mXdXmXdX
697
$endgroup$
add a comment |
$begingroup$
I'm trying to complete this problem. It's been years that I studied probability, so if someone could point me in the right direction, I'd appreciate it.
Let $Z_1, Z_2, ...$ be a sequence of IID random variables with mean $0$ and variance $1$ and define
begin{align*}
S_n = frac{1}{sqrt{n}}sum_{i=1}^{n}Z_i
end{align*}
and
begin{align*}
L_m=lim_{n to infty}E[S_{n}^{m}].
end{align*}
Another method of proof of the Central Limit theorem works by showing that for each $m$, the limit $L_m$ exists, and the sequence satisfies the recurrence relation
begin{align*}
L_{m+1}=mL_{m-1}.
end{align*}
1) Use integration by parts to show that the sequence $R_m=E[Z^{m}]$, where $Z$ is a normal $N(0,1)$ random variable, satisfies the same recurrence relation.
2) Use induction to deduce that the above recurrence relation implies $L_m=R_m=0$ if $m$ is odd, and that they are both equal to
begin{align*}
frac{m!}{2^{m/2}(m/2)!}
end{align*}
if $m$ is even.
Any guidance at all would be appreciated. I'm not sure where to begin here.
probability-theory
asked Jan 12 at 14:22
mXdXmXdX
697
$endgroup$
add a comment |
$begingroup$
I'm trying to complete this problem. It's been years that I studied probability, so if someone could point me in the right direction, I'd appreciate it.
Let $Z_1, Z_2, ...$ be a sequence of IID random variables with mean $0$ and variance $1$ and define
begin{align*}
S_n = frac{1}{sqrt{n}}sum_{i=1}^{n}Z_i
end{align*}
and
begin{align*}
L_m=lim_{n to infty}E[S_{n}^{m}].
end{align*}
Another method of proof of the Central Limit theorem works by showing that for each $m$, the limit $L_m$ exists, and the sequence satisfies the recurrence relation
begin{align*}
L_{m+1}=mL_{m-1}.
end{align*}
1) Use integration by parts to show that the sequence $R_m=E[Z^{m}]$, where $Z$ is a normal $N(0,1)$ random variable, satisfies the same recurrence relation.
2) Use induction to deduce that the above recurrence relation implies $L_m=R_m=0$ if $m$ is odd, and that they are both equal to
begin{align*}
frac{m!}{2^{m/2}(m/2)!}
end{align*}
if $m$ is even.
Any guidance at all would be appreciated. I'm not sure where to begin here.
probability-theory
asked Jan 12 at 14:22
mXdXmXdX
697
$endgroup$
I'm trying to complete this problem. It's been years that I studied probability, so if someone could point me in the right direction, I'd appreciate it.
Let $Z_1, Z_2, ...$ be a sequence of IID random variables with mean $0$ and variance $1$ and define
begin{align*}
S_n = frac{1}{sqrt{n}}sum_{i=1}^{n}Z_i
end{align*}
and
begin{align*}
L_m=lim_{n to infty}E[S_{n}^{m}].
end{align*}
Another method of proof of the Central Limit theorem works by showing that for each $m$, the limit $L_m$ exists, and the sequence satisfies the recurrence relation
begin{align*}
L_{m+1}=mL_{m-1}.
end{align*}
1) Use integration by parts to show that the sequence $R_m=E[Z^{m}]$, where $Z$ is a normal $N(0,1)$ random variable, satisfies the same recurrence relation.
2) Use induction to deduce that the above recurrence relation implies $L_m=R_m=0$ if $m$ is odd, and that they are both equal to
begin{align*}
frac{m!}{2^{m/2}(m/2)!}
end{align*}
if $m$ is even.
Any guidance at all would be appreciated. I'm not sure where to begin here.
probability-theory
probability-theory
asked Jan 12 at 14:22
mXdXmXdX
697
asked Jan 12 at 14:22
mXdXmXdX
697
asked Jan 12 at 14:22
mXdXmXdX
697
asked Jan 12 at 14:22
mXdXmXdX
697
asked Jan 12 at 14:22
mXdXmXdX
697
697
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
1) Write $u=x^m,,v=-frac{1}{sqrt{2pi}}exp-frac{x^2}{2}$ so $lim_{xtopminfty}uv=0$ and $$R_{m+1}=int_{-infty}^infty uv^prime dx=-int_{-infty}^infty u^prime vdx=mR_{m-1}.$$
2) Since $E[S_n^0]=1$ and $E[S_n^1]=0$ for any $n$, $L_m=R_m$ for $min{0,,1}$. Thus $L_m=R_m$ for all $m$. The proof by induction that $L_m=0$ for $m$ odd is trivial; for $m$ even it suffices to note $frac{0!}{2^00!}=1=L_2$, and $$frac{(m+1)(m+2)}{2cdot (m/2+1)}=m+1=frac{L_{m+2}}{m}$$for even $mge 0$.
answered Jan 12 at 14:46
J.G.J.G.
26k22539
$endgroup$
$begingroup$
Thanks! Would you mind explaining a bit how you chose your u, v for the integration?
$endgroup$
– mXdX
Jan 12 at 14:55
1
$begingroup$
@mXdX Given that writing the statement $R_{m+1}=mR_{m-1}$ as integrals gives $int_0^inftyfrac{x^{m+1}}{sqrt{2pi}}exp-frac{x^2}{2}dx=int_0^inftyfrac{mx^{m-1}}{sqrt{2pi}}exp-frac{x^2}{2}dx$, you can't help but notice the relevance of $mx^{m-1}=(x^m)^prime,,(exp-x^2/2)^prime=-xexp-x^2/2$.
$endgroup$
– J.G.
Jan 12 at 15:06
1
$begingroup$
@mXdX Well, in general: given a recursion relation expressing the $n$th term in terms of (sum or all of) the previous $k$ terms, the first $k$ terms determine the sequence. See if you can prove by induction that sequences agreeing in the first terms, as well as on the recurrence relation, must be identical. (In the case at hand, $k=2$, but only the second most recent prior term is needed.)
$endgroup$
– J.G.
Jan 12 at 18:40
1
$begingroup$
@mXdX Thanks; fixed.
$endgroup$
– J.G.
Jan 12 at 18:53
1
$begingroup$
$S_n^0=1,,S_n^1=S_n$.
$endgroup$
– J.G.
Jan 12 at 20:02
|
show 8 more comments
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$begingroup$
1) Write $u=x^m,,v=-frac{1}{sqrt{2pi}}exp-frac{x^2}{2}$ so $lim_{xtopminfty}uv=0$ and $$R_{m+1}=int_{-infty}^infty uv^prime dx=-int_{-infty}^infty u^prime vdx=mR_{m-1}.$$
2) Since $E[S_n^0]=1$ and $E[S_n^1]=0$ for any $n$, $L_m=R_m$ for $min{0,,1}$. Thus $L_m=R_m$ for all $m$. The proof by induction that $L_m=0$ for $m$ odd is trivial; for $m$ even it suffices to note $frac{0!}{2^00!}=1=L_2$, and $$frac{(m+1)(m+2)}{2cdot (m/2+1)}=m+1=frac{L_{m+2}}{m}$$for even $mge 0$.
answered Jan 12 at 14:46
J.G.J.G.
26k22539
$endgroup$
$begingroup$
Thanks! Would you mind explaining a bit how you chose your u, v for the integration?
$endgroup$
– mXdX
Jan 12 at 14:55
1
$begingroup$
@mXdX Given that writing the statement $R_{m+1}=mR_{m-1}$ as integrals gives $int_0^inftyfrac{x^{m+1}}{sqrt{2pi}}exp-frac{x^2}{2}dx=int_0^inftyfrac{mx^{m-1}}{sqrt{2pi}}exp-frac{x^2}{2}dx$, you can't help but notice the relevance of $mx^{m-1}=(x^m)^prime,,(exp-x^2/2)^prime=-xexp-x^2/2$.
$endgroup$
– J.G.
Jan 12 at 15:06
1
$begingroup$
@mXdX Well, in general: given a recursion relation expressing the $n$th term in terms of (sum or all of) the previous $k$ terms, the first $k$ terms determine the sequence. See if you can prove by induction that sequences agreeing in the first terms, as well as on the recurrence relation, must be identical. (In the case at hand, $k=2$, but only the second most recent prior term is needed.)
$endgroup$
– J.G.
Jan 12 at 18:40
1
$begingroup$
@mXdX Thanks; fixed.
$endgroup$
– J.G.
Jan 12 at 18:53
1
$begingroup$
$S_n^0=1,,S_n^1=S_n$.
$endgroup$
– J.G.
Jan 12 at 20:02
|
show 8 more comments
$begingroup$
1) Write $u=x^m,,v=-frac{1}{sqrt{2pi}}exp-frac{x^2}{2}$ so $lim_{xtopminfty}uv=0$ and $$R_{m+1}=int_{-infty}^infty uv^prime dx=-int_{-infty}^infty u^prime vdx=mR_{m-1}.$$
2) Since $E[S_n^0]=1$ and $E[S_n^1]=0$ for any $n$, $L_m=R_m$ for $min{0,,1}$. Thus $L_m=R_m$ for all $m$. The proof by induction that $L_m=0$ for $m$ odd is trivial; for $m$ even it suffices to note $frac{0!}{2^00!}=1=L_2$, and $$frac{(m+1)(m+2)}{2cdot (m/2+1)}=m+1=frac{L_{m+2}}{m}$$for even $mge 0$.
answered Jan 12 at 14:46
J.G.J.G.
26k22539
$endgroup$
$begingroup$
Thanks! Would you mind explaining a bit how you chose your u, v for the integration?
$endgroup$
– mXdX
Jan 12 at 14:55
1
$begingroup$
@mXdX Given that writing the statement $R_{m+1}=mR_{m-1}$ as integrals gives $int_0^inftyfrac{x^{m+1}}{sqrt{2pi}}exp-frac{x^2}{2}dx=int_0^inftyfrac{mx^{m-1}}{sqrt{2pi}}exp-frac{x^2}{2}dx$, you can't help but notice the relevance of $mx^{m-1}=(x^m)^prime,,(exp-x^2/2)^prime=-xexp-x^2/2$.
$endgroup$
– J.G.
Jan 12 at 15:06
1
$begingroup$
@mXdX Well, in general: given a recursion relation expressing the $n$th term in terms of (sum or all of) the previous $k$ terms, the first $k$ terms determine the sequence. See if you can prove by induction that sequences agreeing in the first terms, as well as on the recurrence relation, must be identical. (In the case at hand, $k=2$, but only the second most recent prior term is needed.)
$endgroup$
– J.G.
Jan 12 at 18:40
1
$begingroup$
@mXdX Thanks; fixed.
$endgroup$
– J.G.
Jan 12 at 18:53
1
$begingroup$
$S_n^0=1,,S_n^1=S_n$.
$endgroup$
– J.G.
Jan 12 at 20:02
|
show 8 more comments
$begingroup$
1) Write $u=x^m,,v=-frac{1}{sqrt{2pi}}exp-frac{x^2}{2}$ so $lim_{xtopminfty}uv=0$ and $$R_{m+1}=int_{-infty}^infty uv^prime dx=-int_{-infty}^infty u^prime vdx=mR_{m-1}.$$
2) Since $E[S_n^0]=1$ and $E[S_n^1]=0$ for any $n$, $L_m=R_m$ for $min{0,,1}$. Thus $L_m=R_m$ for all $m$. The proof by induction that $L_m=0$ for $m$ odd is trivial; for $m$ even it suffices to note $frac{0!}{2^00!}=1=L_2$, and $$frac{(m+1)(m+2)}{2cdot (m/2+1)}=m+1=frac{L_{m+2}}{m}$$for even $mge 0$.
answered Jan 12 at 14:46
J.G.J.G.
26k22539
$endgroup$
1) Write $u=x^m,,v=-frac{1}{sqrt{2pi}}exp-frac{x^2}{2}$ so $lim_{xtopminfty}uv=0$ and $$R_{m+1}=int_{-infty}^infty uv^prime dx=-int_{-infty}^infty u^prime vdx=mR_{m-1}.$$
2) Since $E[S_n^0]=1$ and $E[S_n^1]=0$ for any $n$, $L_m=R_m$ for $min{0,,1}$. Thus $L_m=R_m$ for all $m$. The proof by induction that $L_m=0$ for $m$ odd is trivial; for $m$ even it suffices to note $frac{0!}{2^00!}=1=L_2$, and $$frac{(m+1)(m+2)}{2cdot (m/2+1)}=m+1=frac{L_{m+2}}{m}$$for even $mge 0$.
answered Jan 12 at 14:46
J.G.J.G.
26k22539
edited Jan 16 at 15:51
answered Jan 12 at 14:46
J.G.J.G.
26k22539
answered Jan 12 at 14:46
J.G.J.G.
26k22539
answered Jan 12 at 14:46
J.G.J.G.
26k22539
26k22539
$begingroup$
Thanks! Would you mind explaining a bit how you chose your u, v for the integration?
$endgroup$
– mXdX
Jan 12 at 14:55
1
$begingroup$
@mXdX Given that writing the statement $R_{m+1}=mR_{m-1}$ as integrals gives $int_0^inftyfrac{x^{m+1}}{sqrt{2pi}}exp-frac{x^2}{2}dx=int_0^inftyfrac{mx^{m-1}}{sqrt{2pi}}exp-frac{x^2}{2}dx$, you can't help but notice the relevance of $mx^{m-1}=(x^m)^prime,,(exp-x^2/2)^prime=-xexp-x^2/2$.
$endgroup$
– J.G.
Jan 12 at 15:06
1
$begingroup$
@mXdX Well, in general: given a recursion relation expressing the $n$th term in terms of (sum or all of) the previous $k$ terms, the first $k$ terms determine the sequence. See if you can prove by induction that sequences agreeing in the first terms, as well as on the recurrence relation, must be identical. (In the case at hand, $k=2$, but only the second most recent prior term is needed.)
$endgroup$
– J.G.
Jan 12 at 18:40
1
$begingroup$
@mXdX Thanks; fixed.
$endgroup$
– J.G.
Jan 12 at 18:53
1
$begingroup$
$S_n^0=1,,S_n^1=S_n$.
$endgroup$
– J.G.
Jan 12 at 20:02
|
show 8 more comments
$begingroup$
Thanks! Would you mind explaining a bit how you chose your u, v for the integration?
$endgroup$
– mXdX
Jan 12 at 14:55
1
$begingroup$
@mXdX Given that writing the statement $R_{m+1}=mR_{m-1}$ as integrals gives $int_0^inftyfrac{x^{m+1}}{sqrt{2pi}}exp-frac{x^2}{2}dx=int_0^inftyfrac{mx^{m-1}}{sqrt{2pi}}exp-frac{x^2}{2}dx$, you can't help but notice the relevance of $mx^{m-1}=(x^m)^prime,,(exp-x^2/2)^prime=-xexp-x^2/2$.
$endgroup$
– J.G.
Jan 12 at 15:06
1
$begingroup$
@mXdX Well, in general: given a recursion relation expressing the $n$th term in terms of (sum or all of) the previous $k$ terms, the first $k$ terms determine the sequence. See if you can prove by induction that sequences agreeing in the first terms, as well as on the recurrence relation, must be identical. (In the case at hand, $k=2$, but only the second most recent prior term is needed.)
$endgroup$
– J.G.
Jan 12 at 18:40
1
$begingroup$
@mXdX Thanks; fixed.
$endgroup$
– J.G.
Jan 12 at 18:53
1
$begingroup$
$S_n^0=1,,S_n^1=S_n$.
$endgroup$
– J.G.
Jan 12 at 20:02
$begingroup$
Thanks! Would you mind explaining a bit how you chose your u, v for the integration?
$endgroup$
– mXdX
Jan 12 at 14:55
$begingroup$
Thanks! Would you mind explaining a bit how you chose your u, v for the integration?
$endgroup$
– mXdX
Jan 12 at 14:55
1
1
$begingroup$
@mXdX Given that writing the statement $R_{m+1}=mR_{m-1}$ as integrals gives $int_0^inftyfrac{x^{m+1}}{sqrt{2pi}}exp-frac{x^2}{2}dx=int_0^inftyfrac{mx^{m-1}}{sqrt{2pi}}exp-frac{x^2}{2}dx$, you can't help but notice the relevance of $mx^{m-1}=(x^m)^prime,,(exp-x^2/2)^prime=-xexp-x^2/2$.
$endgroup$
– J.G.
Jan 12 at 15:06
$begingroup$
@mXdX Given that writing the statement $R_{m+1}=mR_{m-1}$ as integrals gives $int_0^inftyfrac{x^{m+1}}{sqrt{2pi}}exp-frac{x^2}{2}dx=int_0^inftyfrac{mx^{m-1}}{sqrt{2pi}}exp-frac{x^2}{2}dx$, you can't help but notice the relevance of $mx^{m-1}=(x^m)^prime,,(exp-x^2/2)^prime=-xexp-x^2/2$.
$endgroup$
– J.G.
Jan 12 at 15:06
1
1
$begingroup$
@mXdX Well, in general: given a recursion relation expressing the $n$th term in terms of (sum or all of) the previous $k$ terms, the first $k$ terms determine the sequence. See if you can prove by induction that sequences agreeing in the first terms, as well as on the recurrence relation, must be identical. (In the case at hand, $k=2$, but only the second most recent prior term is needed.)
$endgroup$
– J.G.
Jan 12 at 18:40
$begingroup$
@mXdX Well, in general: given a recursion relation expressing the $n$th term in terms of (sum or all of) the previous $k$ terms, the first $k$ terms determine the sequence. See if you can prove by induction that sequences agreeing in the first terms, as well as on the recurrence relation, must be identical. (In the case at hand, $k=2$, but only the second most recent prior term is needed.)
$endgroup$
– J.G.
Jan 12 at 18:40
1
1
$begingroup$
@mXdX Thanks; fixed.
$endgroup$
– J.G.
Jan 12 at 18:53
$begingroup$
@mXdX Thanks; fixed.
$endgroup$
– J.G.
Jan 12 at 18:53
1
1
$begingroup$
$S_n^0=1,,S_n^1=S_n$.
$endgroup$
– J.G.
Jan 12 at 20:02
$begingroup$
$S_n^0=1,,S_n^1=S_n$.
$endgroup$
– J.G.
Jan 12 at 20:02
|
show 8 more comments
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