Variance and Standard Deviation

The mean \(\mu\) is defined as the center of mass of the distribution, the variance \(\sigma^2\) is the average of squared differences from the mean \(\mu\), and the standard deviation \(\sigma\) tells us how spread out the distribution is.

If we apply a function \(g(\cdot)\) to \(\mathbf{X}\), with \(g(\mathbf{X}) = \mathbf{X}^2\), then:

\[ \begin{aligned} \mathbf{Y} &= \mathbf{X}^2 \\ E(\mathbf{Y}) &= \int_{-\infty}^{\infty} x^2 f(x) \, dx \end{aligned} \]

If \(\mathbf{X}\) and \(\mathbf{Y}\) are independent, then:

\[ \text{Var}(\mathbf{X} + \mathbf{Y}) = \text{Var}(\mathbf{X}) + \text{Var}(\mathbf{Y}) \]

Let the random variable \(\mathbf{X}\) follow a normal distribution:

\[ \mathbf{X} \sim \mathcal{N}(\mu, \sigma^2) \]

Then, the PDF of the normal distribution is:

\[ f(x) = \frac{1}{\sqrt{2\pi} \sigma} e^{-\frac{(x - \mu)^2}{2 \sigma^2}} \]

where \(\mu\) is \(E(\mathbf{X})\) and \(\sigma^2\) is the variance of \(\mathbf{X}\).

To compute \(\text{Var}(\mathbf{X})\), we have:

\[ \begin{aligned} \text{Var}(\mathbf{X}) &= E((\mathbf{X} - \mu)^2) \\ &= E(\mathbf{X}^2) - [E(\mathbf{X})]^2 \end{aligned} \]

Proof:

\[ \begin{aligned} \text{Var}(\mathbf{X}) &= E((\mathbf{X} - \mu)^2) \\ &= E(\mathbf{X}^2 - 2\mu \mathbf{X} + \mu^2) \\ &= E(\mathbf{X}^2) - E(2\mu \mathbf{X}) + E(\mu^2) \\ &= E(\mathbf{X}^2) - 2\mu E(\mathbf{X}) + \mu^2 \\ &= E(\mathbf{X}^2) - \mu^2 \end{aligned} \]

Example:

Given \(T \sim \text{Exp}(\lambda)\), where the PDF is \(f(t) = \lambda e^{-\lambda t}\), the mean is:

\[ \begin{aligned} E(T) &= \int_0^\infty t f(t) \, dt \\ &= \int_0^\infty t \lambda e^{-\lambda t} \, dt \\ &= \left[ -t e^{-\lambda t} \right]_0^\infty + \int_0^\infty e^{-\lambda t} \, dt \\ &= 0 - \frac{1}{\lambda} \left[ e^{-\lambda t} \right]_0^\infty \\ &= \frac{1}{\lambda} \end{aligned} \]

Variance:

\[ \begin{aligned} \text{Var}(T) &= E(T^2) - [E(T)]^2 \end{aligned} \]

To compute \(E(T^2)\), we have:

\[ \begin{aligned} E(T^2) &= \int_0^\infty t^2 \lambda e^{-\lambda t} \, dt \\ &= \frac{2}{\lambda^2} \end{aligned} \]

Then:

\[ \begin{aligned} \text{Var}(T) &= \frac{2}{\lambda^2} - \frac{1}{\lambda^2} \\ &= \frac{1}{\lambda^2} \end{aligned} \]

Standard Deviation:

\[ \sigma(T) = \sqrt{\frac{1}{\lambda^2}} = \frac{1}{\lambda} \]

Median:

\[ \begin{aligned} F(t) &= \int_0^t f(t) \, dt \\ &= 1 - e^{-\lambda t} = \frac{1}{2} \end{aligned} \]

Then, we solve \(e^{-\lambda t} = \frac{1}{2}\), which gives:

\[ t = \frac{\ln(2)}{\lambda} \]

Thus, the median is:

\[ \text{Median}(T) = \frac{\ln(2)}{\lambda} \]