Out of the box & WOW effect Probability

Question

It is a common knowledge, that problems from graph theory/combinatorics can be often solved via short and astonishing probabilistic reasoning. In contrast to this, it is much harder to find such "out of the box"/"WOW effect" solutions for analysis, topology or algebra problems. So, what are your favorite problems like this, how do you prove them?

This question was inspired by a very nice analysis problem:

evaluate $ \ \lim_{n\to \infty} e^{-n} \cdot \sum_{k=0}^{n}\frac{n^k}{k!}$.

with the following, extremely beautiful solution:

This is $P[N_n\leqslant n]$ where $N_n$ is a random variable with Poisson distribution of parameter $n$. Hence each $N_n$ is distributed like $X_1+\cdots+X_n$ where the random variables $(X_k)$ are independent and identically distributed with Poisson distribution of parameter $1$.

By the central limit theorem, $Y_n=\frac1{\sqrt{n}}(X_1+\cdots+X_n-n)$ converges in distribution to a standard normal random variable $Z$, in particular, $P[Y_n\leqslant 0]\to P[Z\leqslant0]$.

Finally, $P[Z\leqslant0]=\frac12$ and $[N_n\leqslant n]=[Y_n\leqslant 0]$ hence $P[N_n\leqslant n]\to\frac12$, QED.

Both, question and solution come from Evaluating $\lim\limits_{n\to\infty} e^{-n} \sum\limits_{k=0}^{n} \frac{n^k}{k!}$.

I think that such a list might be helpful for educational purposes. Such problems act on the imagination of students/listeners and much more likely raise their interest.

Answer 1

Here are two nice (and somewhat related) applications of probabilistic proofs. They require more sophisticated machinery than your example though.

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Fundamental Theorem of Algebra: Every non constant polynomial $p\in\mathbb C[X]$ has a root in $\mathbb C$.

Consider a Brownian motion $B_t$ on $\mathbb C$, and suppose $p$ had no complex roots. Then $f=\frac{1}{p}$ is analytic and bounded. As $f$ is continous and non-constant, we can find reals $\alpha<\beta$ such that $\{\Re f(z)\leq\alpha\}$ and $\{\Re f(z)\geq\beta\}$ both contain (disjoint) open discs.

Then $\Re f(B_t)$ is a bounded martingale, so by the martingale convergence theorem, it converges almost surely. However, Brownian motion is recurrent on $\mathbb C$, so $B_t$ visits $\{\Re f(z)\leq\alpha\}$ and $\{\Re f(z)\geq\beta\}$ infinitely many times. So $$\lim\inf\Re f(z)\leq\alpha<\beta\leq\lim\sup\Re f(z),$$ which contradicts martingale convergence.

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If $f:\mathbb Z^2\to[0,\infty)$ satisfies $$f(x,y)=\frac{f(x+1, y)+f(x, y+1) + f(x-1, y) +f(x, y-1)}{4}$$ for all $x,y\in\mathbb Z$, then $f$ is constant.

Let $X_n$ be a simple symmetric random walk on $\mathbb Z^2$. The condition implies that $M_n=f(X_n)$ is martingale. As $f\geq0$, we know by the martingale convergence theorem that $M_n$ converges almost surely.

But $(X_n)$ is irreducible and recurrent, so (almost surely) visits every point in $\mathbb Z^2$ infinitely often. So $M_n$ hits every value in the image of $f$ infinitely often. This contradicts the convergence of $M_n$ unless $f$ is constant.

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Away from probabilistic arguments, I will always be partial to the Gaussian integral. I particularly like two proofs: the first is the standard change of variables to polars. The second is applying the Feynman trick to $$F(t)=\left(\int_0^te^{-x^2}\mathop{}\!\mathrm{d}x\right)^2.$$ I also like Axler's proof that every linear operator $T$ on a finite-dimensional vector space $V$ over $\mathbb C$ has an eigenvalue. He doesn't define need to define determinants/characteristic polynomials, and instead just considers $v, Tv,\dots, T^nv$, where $n=\dim V$. These vectors are linearly independent, so take a linear combination that equals $0$, and you are done by the fundamental theorem of algebra.

Answer 2

I have lately found a nice problem that matches the description:

Consider $m,n \in \mathbb{Z}_+$ and $p,q\ge 0$ with $p+q=1$. Show that $$(1-p^n)^m+(1-q^m)^n \ \ge 1.$$

It is quite elementary, which makes it actually even more fun.