Hi everybody,

I am working on the cute problem below.

Let X1,..,Xn iid Normal(theta,1) where it is known that theta must be
nonnegative. It is easy to show that
The MLE is theta_MLE=Xbar, if Xbar>=0
The MLE is theta_MLE=0, if Xbar<0

Now the first question of interest is about the asymptotic
distribution of theta_MLE if theta>0. I am trying to show that if
theta>0,
sqrt(n){ theta_MLE - theta } goes in law to Normal(0,1) (equation 1)

I know that sqrt(n){ Xbar - theta } goes in law to Normal(0,1) by the
CLT.
So, I was thinking that it should be enough to show that
P{Xbar<0} goes to 0 when n is large. So I wrote a small proof below.

Xbar -> Normal(theta>0, 1/n) in law
sqrt(n){Xbar -theta} -> Normal(0, 1) in law
P{Xbar<0}=
P{Xbar-theta<-theta}=
P{sqrt(n){Xbar-theta}<-sqrt(n)theta}->Phi(-infty)=0 when n is large

Because P{Xbar<0} goes to 0 when n is large, theta_MLE and Xbar are
asymptotically equal. And so theta_MLE is asymptotically normal as in
(equation 1).

The second question of interest is about the asymptotic distribution
of theta_MLE if theta=0.

What is P{Xbar<0}?
Xbar -> Normal(0, 1/n) in law
sqrt(n){Xbar} -> Normal(0, 1)
P{Xbar<0}=P{sqrt(n){Xbar}<0}->Phi(0)=1/2
And when Xbar<0, theta_MLE=0.

Thus P{Xbar>=0}=1/2. And when Xbar>=0, theta_MLE=Xbar. By the CLT,
when Xbar>=0,
sqrt(n){ theta_MLE - theta } goes in law to Normal(0,1) (equation 1)

I greatly appreciate your comments and help.

lydiajone...@aim.com wrote:
Hi everybody,

I am working on the cute problem below.

Let X1,..,Xn iid Normal(theta,1) where it is known that theta must be
nonnegative. It is easy to show that
The MLE is theta_MLE=Xbar, if Xbar>=0
The MLE is theta_MLE=0, if Xbar<0

Now the first question of interest is about the asymptotic
distribution of theta_MLE if theta>0. I am trying to show that if
theta>0,
sqrt(n){ theta_MLE - theta } goes in law to Normal(0,1) (equation 1)

I know that sqrt(n){ Xbar - theta } goes in law to Normal(0,1) by the
CLT.
So, I was thinking that it should be enough to show that
P{Xbar<0} goes to 0 when n is large. So I wrote a small proof below.

Xbar -> Normal(theta>0, 1/n) in law
sqrt(n){Xbar -theta} -> Normal(0, 1) in law
P{Xbar<0}=
P{Xbar-theta<-theta}=
P{sqrt(n){Xbar-theta}<-sqrt(n)theta}->Phi(-infty)=0 when n is large

Because P{Xbar<0} goes to 0 when n is large, theta_MLE and Xbar are
asymptotically equal. And so theta_MLE is asymptotically normal as in
(equation 1).

The second question of interest is about the asymptotic distribution
of theta_MLE if theta=0.

What is P{Xbar<0}?
Xbar -> Normal(0, 1/n) in law
sqrt(n){Xbar} -> Normal(0, 1)
P{Xbar<0}=P{sqrt(n){Xbar}<0}->Phi(0)=1/2
And when Xbar<0, theta_MLE=0.

Thus P{Xbar>=0}=1/2. And when Xbar>=0, theta_MLE=Xbar. By the CLT,
when Xbar>=0,
sqrt(n){ theta_MLE - theta } goes in law to Normal(0,1) (equation 1)

I greatly appreciate your comments and help.

I don't think you need to split the proof into two cases. I do think
you need to be a bit careful about what you mean by convergence in law.
Saying that sqrt(n)*theta_MLE is asymptotically normal is not exactly
the same thing as saying that theta_MLE is asymptotically normal.

Let Fn be the CDF of theta_MLE and Phi the standard normal CDF. Without
regard to theta being positive v. zero, we know that for positive values
of a, theta_MLE < a iff Xbar < a, and for nonpositive values of a we
know that theta_MLE < a cannot occur. So Fn(a) = Phi(sqrt(n)(a -
theta)) if a > 0, Fn(a) = 0 otherwise. Therefore

sup_a |Fn(a) - Phi(sqrt(n)(a - theta))| = sup_{a<=0} Phi(sqrt(n)(a -
theta)) = Phi(-sqrt(n)theta) -> 0 as n -> infinity.

So while I hesitate to phrase it as convergence in law (theta_MLE is
really converging to a constant), you can argue that for large n the
distribution of theta_MLE is uniformly close to a normal distribution,
without worrying about theta > 0 v. theta = 0.
Thanks again.

/Paul