Variance of linear combos of R.Vs, Markov's and Chebyshev's inequalities

A hw problem on variance

Continuing to chug my way through homework problems, I worked through this problem which asks


$$ f_{X,Y}(x,y) = \begin{cases} c(x+y) & 0 \leq x \leq 1, 0 \leq y \leq 2 \\ 0 & \text{otherwise} \end{cases} $$

Where $c$ is a constant. Find $V(2X - 3Y + 8)$.

This was a pretty straightforward problem, albeit a tad tedious as in taking all of these steps

  1. Finding the constant c by noting that integrating the joint distribution must equal 1
  2. Finding the marginal distributions for X and Y
  3. Finding E(X), E(Y), V(X), V(Y) and Cov(X,Y)
  4. Breaking down V(2X - 3Y +8) into components of the above to make the final calculation

It still helped to play with the formulas and iron in the concepts of expectation and variance. It also was a good refresher on finding marginal distributions by integrating out variables within a joint distribution.

Probability inequalities

I got started reading about probability inequalities. The first two theorems, Markov's inequality and Chebyshev's inequality are pretty easy to understand. As the book puts it

Inequalities are useful for bounding quantities that might otherwise be hard to compute. They will be used in the theory of convergence in the next chapter.

Markov's inequality

Markov's inequality helps bound non-negative random variables. Let's say $X$ is a non-negative random variable, and we know its mean, $E(X)$. We wish to know $P(X > t)$, but let's say $X$'s distribution function is hard to compute for some reason. We can find a bound on $P(X > t)$ by starting with the definition of expectation (note we'll use shorthand $f(x)$ for $f_X(x)$):

$E(X) = \int_0^{\infty} x f(x)dx$

and breaking this into two parts:

$\int_0^{\infty} x f(x)dx = \int_0^{t} x f(x) dx + \int_t^{\infty} x f(x)dx$

which must be greater than

$\int_t^{\infty} x f(x)dx$

from here it follows that

$\int_t^{\infty} x f(x)dx \geq \int_t^{\infty} t f(x)dx \geq t \int_t^{\infty} f(x)dx = t P(X > t)$

So we have

$E(X) \geq t P(X > t)$

which can be rearranged as

$P(X > t) \leq \frac{E(X)}{t}$

and there you have Markov's inequality.

Chebyshev's inequality

We can use this result to reason about $P(|X - \mu| \geq t)$ by plugging this into Markov's inequality, squaring both sides, and noting the definition of variance:

$P(|X-\mu| \geq t) = P(|X-\mu|^2 \geq t^2) \leq \frac{E(X-\mu)^2}{t^2} = \frac{\sigma^2}{t^2}$.

It's pretty neat that such simple derivations lead to results that are so useful. As Wikipedia notes

Chebyshev's inequality guarantees that, for a wide class of probability distributions, "nearly all" values are close to the mean—the precise statement being that no more than $\frac{1}{k^2}$ of the distribution's values can be more than k standard deviations away from the mean... The inequality has great utility because it can be applied to any probability distribution in which the mean and variance are defined.

And to paraphrase wikipedia further, it also provides a more generalized, albeit looser, rule of thumb to the 68–95–99.7 rule which applies to Normal distributions, as under Chebyshev's inequality a minimum of just 75% of values must lie within two standard deviations of the mean and 89% within three standard deviations.

Side project

I continue to work on the automatic notebook generator. Will report more when I have a decent chunk of progress to coherently describe.