A continuous random variable \(X\) takes a range of values, which may be finite or infinite in extent (e.g. \([0, 1]\), \([0, \infty)\), \((−\infty, \infty)\)), whereby the probability of \(X\) falling between two values a and b is:
\[P(a<X<b)=\int_a^b f(x)dx\] For a special given function called a probability distribution function.
A probability distribution function (PDF) is a function \(f(x)\) that satisfies the two properties:
Show the following function is a probability distribution function. \[ \begin{equation*} g(x)=\begin{cases} 0 \quad &\text{if} \, x < 0 \\ 4x^3 \quad &\text{if } \, 0 \leq x \leq 1 \\ 0 \quad &\text{if } \, x > 1 \\ \end{cases} \end{equation*} \]
Show the following function is a probability distribution function. \[ \begin{equation*} g(x)=\begin{cases} 0 \quad &\text{if } \, x < 0 \\ ke^{-kx} \quad &\text{if } \, x \geq 0 \\ \end{cases} \end{equation*} \]
The distribution in (2) is called the exponential distribution.
The normal distribution is one of the most famous distributions:
\[f(x) = \frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{1}{2}(\frac{x-\mu}{\sigma})^2}\]
The parameters \(\mu\) and \(\sigma\) are parameters mean and standard deviation respectively. In this case we represent this as \(X \sim N(\mu,\sigma)\).
If \(Z \sim N(0,1)\), we say that \(Z\) is standard normal. In this case, plugging in \(\mu=0\) and \(\sigma=1\) gives:
\[f(z) = \frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}z^2}\]
The expected value of a random variable \(X\) with probability distribution \(f(x)\) is:
\[E(X)=\int_{-\infty}^\infty xf(x)dx\]
If X is a continuous random variable,
(Hint: Let \(Z = \frac{X-\mu}{\sigma}\), take expected value of both sides, and reduce.)
If \(X\) is a continuous random variable with mean \(\mu\), the variance \(X\) is:
\[Var(X)=E\big((X-\mu)^2\big)\]
(Hint: use integration by parts with \(u=z\) and \(dv=ze^{-z^2}/2dz\)).
If X and Y are independent random variables, then:
Find the mean and variance of the exponential distribution defined in problem 2.