Good evening,

As I keep practicing towards the Level I exam, I want to finish my review of the Quantitative Finance material. This post will hence be the follow up of my previous post. I will here discuss how you test a hypothesis on some statistical measure.

## General Concept

The main concept is as follows, you make an initial hypothesis which is called the *null hypothesis*, $H_0$, and which is the statement you want to reject. If $H_0$ is rejected, we hence can accept the alternative hypothesis $H_a$. Of course in statistics, you can never be sure of anything. Hence, you can only reject the null hypothesis with a certain confidence level $\alpha$. It is important to understand that if you can’t reject the null hypothesis, it does not mean that you can accept it! Hence, rejecting the null hypothesis is more powerful than failing to reject it. So, if you want to prove some statement, you should test for its opposite as the null hypothesis; if you can reject it, then you can accept the alternative hypothesis which is your original statement. A test can either be one-tailed or two-tailed, depending on what you want to test.

## First example

Let’s take a simple example: you measure a sample of the returns of the S&P that you assume to be normal. You measure a sample volatility $\mu_s$ and a sample standard deviation $s$. By the central limit theorem, we now that the estimate of the mean follows a law $\mathcal{N}(\mu_0, \frac{\sigma^2}{n})$ where $\mu_0$ is the population mean and $\sigma$ is the population standard deviation. Now, you cannot use your measure $\mu_s$ to say that you found the true population mean, because it’s just a sample statistic. However, what you can say is that, given the fact that you found the sample statistic $\mu_s$ and a given confidence interval $\alpha$, it is highly improbable that the population mean $\mu_0$ was equal to some value $x$. Hence the null hypothesis is $\mu_0=x$.

Let’s assume that you found that $\mu_s=0.1$, $s=0.25\%$ and $n=250$. You want to show that $\mu_0 \neq 0$, so you null hypothesis is $\mu_0 = 0$ and with a significance level $\alpha=5\%$. You can compute the z-statistic as follows:

$$ z= \frac{\mu_s-\mu_0}{\frac{s}{\sqrt{n}}} \sim \mathcal{N}(0,1) $$

Now if $\mu_0$ was equal to 0, you know that there is 5% chance that the estimate $\mu_s$ would lie outside the range $\mu_0 \pm z_{2.5\%} = 0 \pm 1.96 = \pm 1.96$ (because the sample statistic can lie on both side of the distribution). The z-statistic was computed at 6.33, which is more than 1.96, so you can reject the null hypothesis $\mu_0=0$ and accept the alternative hypothesis $\mu \neq 0$.

This was a two-tailed hypothesis. One-tailed hypothesis would be looking at only one side of the distribution: $H_0 = \mu_0 \geq 0$ or $H_0 = \mu_0 \leq 0$. As a rule of thumb, you can remember the the null hypothesis **always** contains the “equal” sign.

## P-values

The p-value is the probability (assuming the null hypothesis is true) to have a sample statistic at least as extreme as the one being measure. You compute it by look at the test statistic (6.33 in the previous example) and you find the probability (using the Z-table) that a test statistic can be above that value (and you multiply it by 2 if it is two-tailed). In this case, the p-value is very close to 0. Now, if the p-value of the test statistic is below the significance level of the test, you can reject the null hypothesis. This is useful if you want to discuss statistics without having to impose a certain significance level $\alpha$ to the reader; you can just display the p-value and let him decide whether it’s good enough or not.

There are some other hypothesis tests presented in the curriculum, but this is the main framework to remember. It’s pretty easy, and it allows you to score a lot of points in the quantitative finance part of the exam.