How to solve this

Question

Find the distribution function for above exercise. Recall that the density functions of above exercise is f (x) =C(4x −2x2), 0<x <2 0, otherwise

Answer 1

Since f(x) is a probability density function, we must have

\int_{0}^{2}f(x)~dx = 1

⇒ \int_{0}^{2}C(4*x-2*x^{2})~dx = 1

⇒ C\int_{0}^{2}(4*x-2*x^{2})~dx = 1

⇒ \int_{0}^{2}(4*x-2*x^{2})~dx = \frac{1}{C}

Next, we find that

\int_{0}^{2}(4*x-2*x^{2})~dx = 4\int_{0}^{2}x~dx - 2\int_{0}^{2}x^{2}~dx

= 4[\frac{1}{2}x^{2}]_{0}^{2}

- 2[\frac{1}{3}x^{3}]_{0}^{2}

= 4[\frac{1}{2}*2^{2} - \frac{1}{2}*0^{2}]

- 2[\frac{1}{3}*2^{3} - \frac{1}{3}*0^{3}]

= 4[\frac{4}{2} - 0] - 2[\frac{8}{3} - 0]

= 4[2] - 2[\frac{8}{3}]

= 8 - \frac{16}{3}

= \frac{24}{3} - \frac{16}{3}

= \frac{8}{3}

We can now deduce from the preceding work that

\frac{8}{3} = \frac{1}{C}

⇒ \frac{1}{C} = \frac{8}{3}

⇒ C = \frac{3}{8}

⇒ f(x) = \frac{3}{8}(4*x-2*x^{2}), 0 < x < 2

0 otherwise

⇒ f(x) = \frac{12}{8}x - \frac{6}{8}x^{2}, 0 < x < 2

0 otherwise

⇒ f(x) = \frac{3}{2}x - \frac{3}{4}x^{2}, 0 < x < 2

0 otherwise

Finally, we obtain the cumulative distribution function by calculating

F(x) = \int_{0}^{x}f(x)~dx

= \int_{0}^{x}[\frac{3}{2}x - \frac{3}{4}x^{2}]~dx

= \frac{3}{2}\int_{0}^{x}x~dx - \frac{3}{4}\int_{0}^{x}x^{2}~dx

= \frac{3}{2}[\frac{1}{2}x^{2}]_{0}^{x} - \frac{3}{4}[\frac{1}{3}x^{3}]_{0}^{x}

= \frac{3}{2}[\frac{1}{2}x^{2} - \frac{1}{2}*0^{2}]

- \frac{3}{4}[\frac{1}{3}x^{3} - \frac{1}{3}*0^{3}]

= \frac{3}{2}[\frac{1}{2}x^{2} - 0] - \frac{3}{4}[\frac{1}{3}x^{3} - 0]

= \frac{3}{2}[\frac{1}{2}x^{2}] - \frac{3}{4}[\frac{1}{3}x^{3}]

= \frac{3}{4}x^{2} - \frac{3}{12}x^{3}

= \frac{3}{4}x^{2} - \frac{1}{4}x^{3}

⇒ F(x) = \frac{3}{4}x^{2} - \frac{1}{4}x^{3}, 0 < x < 2

1 x > 2