Cumulative distributions

A continuous random variable X that can assume values between 0 and 1 has a probability density function given by:
\[ \begin{align} f(x) &= 2 (1 - x)\quad 0 \le x \le 1 \\ &= 0 \quad \text{elsewhere} \end{align}\] (a) Find the corresponding cumulative density function.
(b) What is the probability of X taking a value less than 0.75?

The cumulative density function is the integral of the probability density function.

\[ \begin{align} \int_{-\infty}^x 2(1-t) dt &= \int_0^x 2(1-t) dt \\ &= \left[ 2t - t^2 \right]_0^x \\ \end{align}\ \] so \[ \begin{align} F(x) &= 2x- x^2 \end{align}\ \]

To find the probability that x is less than 0.75, use the definition of F

\[ \begin{align} P(0 \lt X \le 0.75) &= F(0.75) - F(0) \\ &= (2(\frac{3}{4}) - (\frac{3}{4})^2) - 0 \\ &= 15/16 \end{align}\ \]

You will gain most benefit if you work through each problem first and then come back and check your work against the solution.

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Now You Try:

A continuous random variable X that can assume values between 0 and 1 has a probability density function given by:
\[ \begin{align} f(x) &= 12x^2 (1 - x)\quad 0 \le x \le 1 \\ &= 0 \quad \text{elsewhere} \end{align}\] (a) Find the corresponding cumulative density function.
(b) What is the probability of X taking a value less than 0.75?