The probabilistic proof of the Weierstrass approximation theorem was to show you how probability theory can be used to prove an important theorem in a different field. We did something similar in Example 4.5 (Erdos probabilistic method in combinatorics) and in Examples Sheet 1, #8. The Wikipedia page on Bernstein polynomials has a list of Berstein polynomials and an animation showing how they are used to approximate a continuous function on $[0,1]$.
I cheated today by using a result from the schedules for Analysis II ("a continuous function on a closed bounded interval is uniformly continuous"). I had thought about giving an easier proof by not appealing to Analysis II, but I decided it would be nicer to show you the proper full result.
The "easy" route would have been to make the stronger assumption that there exists $\lambda$ such that $|f(x)-f(y)|\leq \lambda |x-y|$ for all $x,y\in[0,1]$. This is Lipschitz continuity. Then the proof could have been done using only Jensen's inequality:
$E\Bigl[|f(X/n)-f(x)|\Bigr] \leq \lambda E\Bigl[|X/n-x|\Bigr] \leq \lambda \sqrt{E\Bigl[(X/n-x)^2\Bigr]}=\lambda\sqrt{\displaystyle\frac{x(1-x)}{n}}\leq \displaystyle\frac{\lambda}{2\sqrt{n}}$
which $\to 0$ as $n\to\infty$. However, I like giving you the stronger result. This needed Chebyshev's inequality and illustrated a nice idea that is frequently used in probability calculations. We partitioned the sample space into $A$ and $A^c$ and reasoned differently in each. For $A$ we had $P(A)$ small and $|f(X/n)-f(x)|\leq 2$, whereas for $A^c$ we had $P(A^c)\leq 1$ and $|f(X/n)-f(x)|$ small.
Benford's law is surprising when first encountered. The fact that the leading digit should be "1" with a frequency of 0.301 (rather than $1/9$) has been used as a test to check for fraud in financial records. The idea is that if a person is making up numbers, he unlikely to arrange that his distribution of first digits agree with Benford's law. You can read more about that, and many other examples of Benford's law, in the Wikipedia article referenced above.
I cheated today by using a result from the schedules for Analysis II ("a continuous function on a closed bounded interval is uniformly continuous"). I had thought about giving an easier proof by not appealing to Analysis II, but I decided it would be nicer to show you the proper full result.
The "easy" route would have been to make the stronger assumption that there exists $\lambda$ such that $|f(x)-f(y)|\leq \lambda |x-y|$ for all $x,y\in[0,1]$. This is Lipschitz continuity. Then the proof could have been done using only Jensen's inequality:
$E\Bigl[|f(X/n)-f(x)|\Bigr] \leq \lambda E\Bigl[|X/n-x|\Bigr] \leq \lambda \sqrt{E\Bigl[(X/n-x)^2\Bigr]}=\lambda\sqrt{\displaystyle\frac{x(1-x)}{n}}\leq \displaystyle\frac{\lambda}{2\sqrt{n}}$
which $\to 0$ as $n\to\infty$. However, I like giving you the stronger result. This needed Chebyshev's inequality and illustrated a nice idea that is frequently used in probability calculations. We partitioned the sample space into $A$ and $A^c$ and reasoned differently in each. For $A$ we had $P(A)$ small and $|f(X/n)-f(x)|\leq 2$, whereas for $A^c$ we had $P(A^c)\leq 1$ and $|f(X/n)-f(x)|$ small.
Benford's law is surprising when first encountered. The fact that the leading digit should be "1" with a frequency of 0.301 (rather than $1/9$) has been used as a test to check for fraud in financial records. The idea is that if a person is making up numbers, he unlikely to arrange that his distribution of first digits agree with Benford's law. You can read more about that, and many other examples of Benford's law, in the Wikipedia article referenced above.
