Example Give reasons why the following are not valid probability density functions.

1. $f(x) = \begin{cases} \frac{1}{4}x & -1 \leq x \leq 2, \\ 0 & \text{otherwise.} \end{cases}$
2. $f(x) = \begin{cases} x^2 & 1 \leq x \leq 3, \\ 0 & \text{otherwise.} \end{cases}$

a) $f(x)\ does\ not\ stay \geq 0$ — not a valid P.D.F.

b)$\int_1^3 x^2 \, dx = \left[\frac{1}{3}x^3 \right]_1^3 = 9 - \frac{1}{3} = 8 \frac{2}{3}$

$\neq 1$, so it is not a valid P.D.F.

Example The continuous random variable $X$ has a probability density function given by:

a) Find the value of $k$.

b) Find $P(1 < X < 2)$

a) (Hint: Use $\int_{-\infty}^{\infty} f(x) \, dx = 1$)

$\int_1^4 kx^3 \, dx = \left[ \frac{kx^4}{4} \right]_1^4 = \frac{64k}{4} - \frac{k}{4} = 1$

$\Rightarrow \frac{255k}{4} = 1 \Rightarrow k = \frac{4}{255}$

b)$P(1 < X < 2) = \int_1^2 \frac{4}{255} x^3 \, dx = \left[ \frac{x^4}{255} \right]_1^2 = \frac{1}{255}(16 - 1) = \frac{15}{255} = \frac{1}{17}$

(Note: $P(X = k) = \int_k^k f(x) \, dx = 0$ (infinitely thin strip))

$P(a \leq X \leq b) \equiv P(a < X \leq b) \equiv P(a < X < b)$

Example The continuous random variable X has a probability density function given by:

Find $F(x)$

$P(X \leq x) = \int_0^x \frac{3}{8}x^2 \, dx = \left[ \frac{x^3}{8} \right]_0^x = \frac{x^3}{8}$

• Only need to go up to the specified domain of the function. Thus:

$F(x) = \begin{cases} 0 & x \leq 0 \\ \frac{x^3}{8} & 0 \leq x \leq 2 \\ 1 & \text{otherwise} \end{cases}$

Example The continuous random variable $X$ has a cumulative distribution function given by:

Find the probability density function, $f(x)$

• (PDF is the differential of CDF)

$F'(x) = \frac{1}{5}(2x) = \frac{2x}{5}$

Thus:

$f(x) = \begin{cases} \frac{2x}{5} & 2 \leq x \leq 3 \\ 0 & \text{otherwise} \end{cases}$