Finding CDF limit

Question

I am tasked with finding the CDF limit associated with the probability measure $\text{N}(0,n)$ as $n \to \infty$. Defining $F_n(u)$ to be the CDF associated with $\text{N}(0,n)$ and $F(u) = 0$, my intuition is telling me that $F_n(u) \to F(u)$, but when I try to prove this I obtain something else entirely.

My first attempt involved using the Dominated Convergence Theorem to get something along the lines of

$$ \lim_{n \to \infty} F_n(u) = \lim_{n \to \infty} \int_{-\infty}^u \frac{1}{\sqrt{2 \pi n}} e^{- \frac{x^2}{2n}} \text{d}x = \int_{-\infty}^u \lim_{n \to \infty} \frac{1}{\sqrt{2 \pi n}} e^{- \frac{x^2}{2n}} \text{d}x $$

However, this turned out to not be be a fruitful approach since I was unable to find an integrable upper-bounding function. My next attempt (and what is currently causing my confusion) is the following:

Suppose $X \sim \text{N}(0,n)$. We may write $X = \sqrt{n}Z$ where $Z \sim \text{N}(0,1)$. We may then write \begin{align*} F_n(u) &= \mathbb{P}(X \leq u) \\ &= \mathbb{P}(\sqrt{n}Z \leq u) \\ &= \mathbb{P}\left( Z \leq \frac{u}{\sqrt{n}} \right)\\ &= \Phi\left(\frac{u}{\sqrt{n}} \right) \end{align*} From this, we have

$$ \lim_{n \to \infty} F_n(u) = \lim_{n \to \infty} \Phi\left(\frac{u}{\sqrt{n}} \right) = \Phi(0) = \frac{1}{2} $$

I am having a hard time believing this result. As $n \to \infty$, the probability measure becomes a measure with infinite variance but still Normal centered around $0$. Is there any (hopefully intuitive) explanation of this result if it is correct, or perhaps my work is flawed? Perhaps I should return to the DCT approach? Any guidance is greatly appreciated.

Answer 1

Your confusion is because the random variables $X_n\sim N(0,1)$ are not converging to any real valued random variable. The correct answer is $ \lim_{n \to \infty} F_n(u) = \lim_{n \to \infty} \Phi\left(\frac{u}{\sqrt{n}} \right) = \Phi(0) = \frac{1}{2} $ for every $u$.

Note that $\sqrt n Z \to \infty$ if $Z>0$ and $\sqrt n Z \to -\infty$ if $Z<0$. $[P(Z=0)=0$ so we can ignore this]. Let this limit be $Y$. [$Y$ takes only the values $\pm \infty$!]. Thus $P(\sqrt n Z \leq u)$ tends to $P(Y\leq u)=P(Y \neq \infty)=P(Z\leq 0)=\frac 1 2 $.

In your first aproach the limit is easily seen to be $\frac 1 2$ if you make the substitution $y=\frac x {\sqrt n}$.

Answer 2

I just wanted to clarify how there is not a way to use DCT as OP wishes.

if it were possible, we could swap integral and limit to obtain

$$ \lim_{n \to \infty} F_n(u) = \int_{-\infty}^{u} \lim_{n \to \infty} f_n \; dx = 0 \ne \frac{1}{2} $$

which we know, a posteriori, it is not the correct answer.

The reason as to why DCT fails is beacuse $\sup_n |f_n|$ is not integrable and so there is no integrable dominating function.

Indeed, by taking the derivative wrt $n$, we see that

$$ \sup_n |f_n| = \sup_n \frac{e^{-\frac{x^2}{2n}}}{\sqrt{2 \pi n}} = \frac{e^{-\frac{1}{2}}}{\sqrt{2 \pi} |x|} $$

which is not integrable.