Question

Let f(x) = cxe−x2 if x ≥ 0 and f(x) = 0 if x < 0.

(a) For what value of c is f a probability density function?

(b) For that value of c, find P(1 < X < 3).

Answer 1

(a) If f(x) is to be a proper density function, then its integral over the given support must evaulate to 1:

`\displaystyle\int_(-\infty)^\infty f(x)\,\mathrm dx = \int_0^\infty cxe^(-x^2)\,\mathrm dx=1`

For the integral, substitute u = x ² and du = 2x dx. Then as x → 0, u → 0; as x → ∞, u → ∞:

`\displaystyle\frac12\int_0^\infty ce^(-u)\,\mathrm du=\frac c2\left(\lim_(u\to\infty)(-e^(-u))-(-1)\right)=1`

which reduces to

c / 2 (0 + 1) = 1 → c = 2

(b) Find the probability P(1 < X < 3) by integrating the density function over [1, 3] (I'll omit the steps because it's the same process as in (a)):

`\displaystyle\int_1^3 2xe^(-x^2)\,\mathrm dx = \boxed{(e^8-1)/(e^9)} \approx 0.3678`