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Andru [333]
2 years ago
13

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

Mathematics
1 answer:
sergij07 [2.7K]2 years ago
5 0

(a) If <em>f(x)</em> 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 <em>u</em> = <em>x</em> ² and d<em>u</em> = 2<em>x</em> d<em>x</em>. Then as <em>x</em> → 0, <em>u</em> → 0; as <em>x</em> → ∞, <em>u</em> → ∞:

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

which reduces to

<em>c</em> / 2 (0 + 1) = 1   →   <em>c</em> = 2

(b) Find the probability P(1 < <em>X </em>< 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{\frac{e^8-1}{e^9}} \approx 0.3678

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