Vtyjkyuk

Suppose X$_n$ folowes $N(mu,frac1n^2)$. Thus

$$ F_X_n = frac nsqrt2piint_-infty^xexp(-frac12n^2y^2)dy= frac nsqrt2piint_-infty^nx exp(-frac12z^2)dz$$.

Why does the integral boundry change?

It follow that (sorry for the format, couldnt figure out how to typeset a piecwise function here):

$$ lim_n->infty F_X_N(x)= 0 text if x<0; frac12 text if x=0;1 text if x>0$$

Note that $lim_n->infty is not a CDF. However, it does hold that

$$ F_n textconverges in distribution to F $$

$$ F(x) = 0 text if x , 0 ; 1 text if x $geq$ 0$$

What is the reasoning behind these conclusions ?

This exmpale is from a chapter on convergence and includes Chebychev`s Inequality, Theorem of Levy-Cramer and Lemma of Slutsky.

There is a mistake in the second expression for $F_X_n$. There is no $n$ in the last expression for it. The correct formula is $F_X_n=frac 1 sqrt 2pi int_-infty^nxexp(-frac 1 2 z^2), dz$. You get this by making the substitution $z=ny$. (As $y$ ranges from $-infty$ to $x$, $z$ ranges from $-infty$ to $nx$). When you take the limit of $F_X_n(x)$ you get integral over the whole line $mathbb R$ if $x>0$, integral over the empty set if $x<0$ and integral over $(-infty, 0)$ if $x=0$. This gives the limit as $1$,$0$ and $frac 1 2$ respectively. This limiting function is not a distribution function because it is not continuous from the right at $0$. However, if you redefine the value at $0$ to be $1$ you will get a distribution function $F$ and $F_X_n to F$ at all points where $F$ is continuous (namely, points other than $0$); Hence $F_X_n$ converges in distribution.