I just worked through the following example from Shreve's Stochastic Calculus for Finance.
For each positive integer $n$, define $f_n$ to be the normal density
with mean $0$ and variance $n$, i.e.,$$f_n(x)=\frac{1}{\sqrt{2n\pi}}e^{-\frac{x^2}{2n}}\text{.}$$
He defines $f(x) = \lim_{n \to \infty} f_n(x)$. Then he poses the problem
(iii) Note that
$$ \lim_{n \to \infty} \int _{-\infty} ^{\infty} f_n(x)\, dx \neq \int_{-\infty}^{\infty}f(x)\,dx\text{.}$$
Explain why this does not violate the Monotone Convergence Theorem,
Theorem 1.4.5.
It's easy to see that these functions aren't monotonically decreasing, so the assumptions for the Monotone Convergence Theorem aren't satisfied. I now want to understand why this example does not violate the Dominated Convergence Theorem. I believe that it suffices to show that
$$\int_{-\infty}^{\infty} \sup_{n \in \mathbb{N}} f_n(x) \, dx \nless \infty \text{.}$$
I have two questions about this.
- Can you confirm, yes or no, that the above (essentially) proves the $f_n$ do not satisfy the assumption of the Dominated Convergence Theorem?
- Do you have an explanation for why the above display is true? A proof is preferable, the details aren't important.
Best Answer
Hint: Let $x>1$. $\sup_n f_n(x)\geq f_{[x^{2}]}(x)$ where $[,]$ is the greatest integer function. $f_{[x^{2}]}(x) \geq \frac 1 {(2\pi)^{1/2} |x|}e^{-x^{2}/2[x^{2}]}$. Note that $x^{2}/[x^{2}] \to 1$ as $x \to \infty$. Conclude that $\int_0^{\infty} \sup_n f_n(x)dx=\infty$.