Student's t-Distribution

The backbone of hypothesis testing with small sample sizes.

The "Small Sample" Distribution

The t-distribution is a type of probability distribution that is similar to the normal distribution but has heavier tails. This means it assigns more probability to extreme values. It is used in place of the normal distribution when you have small sample sizes (typically n < 30) and the population standard deviation is unknown.

In finance, this is incredibly common. You rarely know the true volatility of an asset and often work with limited historical data. The t-distribution provides a more cautious and robust framework for constructing confidence intervals and performing hypothesis tests (like the t-test) in these real-world scenarios.

Interactive t-Distribution

Adjust the degrees of freedom to see how the t-distribution compares to the standard normal distribution. Notice how it converges to the normal distribution as `df` increases.

Degrees of Freedom (df): 5

Mean (

\mu

): 0.00

Variance (

\sigma^2

): 1.67

Core Concepts

Probability Density Function (PDF)

The PDF is defined by its single parameter: the degrees of freedom (

\nu

or `df`).

f(t) = \frac{\Gamma(\frac{\nu+1}{2})}{\sqrt{\nu\pi}\Gamma(\frac{\nu}{2})} \left(1 + \frac{t^2}{\nu}\right)^{-\frac{\nu+1}{2}}

$\nu$ (nu) represents the degrees of freedom, which is typically the sample size minus one (n - 1).
$\Gamma$ is the Gamma function.

Expected Value & Variance

Expected Value (Mean)

E[X] = 0 \quad \text{for } \nu > 1

Variance

Var(X) = \frac{\nu}{\nu - 2} \quad \text{for } \nu > 2

Notice that the variance is always greater than 1, reflecting the "fatter tails" compared to the standard normal distribution (which has a variance of 1).

Key Derivations

Deriving the Moments

The mean and variance are derived by integrating over the PDF. These derivations show why the degrees of freedom are critical.

Deriving the Expected Value (Mean)

Step 1: Set up the Integral for E[T]

The expected value is the integral of $t \cdot f(t)$ over its domain $(-\infty, \infty)$ .

E[T] = \int_{-\infty}^{\infty} t \cdot \frac{\Gamma(\frac{\nu+1}{2})}{\sqrt{\nu\pi}\Gamma(\frac{\nu}{2})} \left(1 + \frac{t^2}{\nu}\right)^{-\frac{\nu+1}{2}} dt

Step 2: Analyze the Integrand

Let's look at the function being integrated. The constant part can be pulled out. The core of the function is $g(t) = t \cdot \left(1 + \frac{t^2}{\nu}\right)^{-\frac{\nu+1}{2}}$ .

We can check if this is an odd or even function. Let's find $g(-t)$ :

g(-t) = (-t) \cdot \left(1 + \frac{(-t)^2}{\nu}\right)^{-\frac{\nu+1}{2}} = -t \cdot \left(1 + \frac{t^2}{\nu}\right)^{-\frac{\nu+1}{2}} = -g(t)

Since $g(-t) = -g(t)$ , the integrand is an **odd function**. The integral of any odd function over a symmetric interval like $(-\infty, \infty)$ is zero.

Step 3: Conclusion (with a condition)

This integral is guaranteed to be zero **if and only if it converges**. The integral converges only when the degrees of freedom $\nu > 1$ .

Final Mean Formula

E[T] = 0 \quad (\text{for } \nu > 1)

Deriving the Variance

We use $Var(T) = E[T^2] - (E[T])^2$ . Since $E[T]=0$ (for ν &gt 2, which is required for variance anyway), we just need to find $E[T^2]$ .

Step 1: Set up the Integral for E[T²]

E[T^2] = \int_{-\infty}^{\infty} t^2 \cdot f(t) dt

The integrand $t^2 \cdot f(t)$ is an **even function**, so we can simplify the integral:

E[T^2] = 2 \int_{0}^{\infty} t^2 \cdot \frac{\Gamma(\frac{\nu+1}{2})}{\sqrt{\nu\pi}\Gamma(\frac{\nu}{2})} \left(1 + \frac{t^2}{\nu}\right)^{-\frac{\nu+1}{2}} dt

Step 2: Use Substitution

This integral is complex. The standard solution involves a substitution related to the Beta function. Let $y = \frac{1}{1+t^2/\nu}$ .

After significant algebraic manipulation and applying the properties of the Beta and Gamma functions, the integral simplifies.

Step 3: The Result

The integral evaluates to $E[T^2] = \frac{\nu}{\nu-2}$ , but this only converges when $\nu > 2$ .

Final Variance Formula

Var(T) = \frac{\nu}{\nu - 2} \quad (\text{for } \nu > 2)