Question

A1. Let X be a continuous r.v with density function
f(x) = ( |x|
10
, for r 2 < x < 4
0, elsewhere
. Calculate the expected value of X.

Answer 1

$f(x)=\alpha$ |x|¹⁰.

As$\int_{0}^{+\infty} f(t) = 1$ we can find $\alpha$.

$\int_{0}^{+\infty} f(x) = \alpha\int_{2}^{4} x^{10}= \alpha \frac{x^{11}}{11}\bigg|^{4}_{2}$

$=\alpha \frac{x^{11}}{11} |_2^4=\frac{11}{4^{11}-2^{11}} \newline \alpha=\frac{11}{4^{11}-2^{11}}$

$EX=\int_{2}^{4}xf(x)=\alpha \int_{2}^{4} x^{11}dx = \newline =\{ \alpha \frac{x^{12}}{12} \} \bigg|_2^4=\alpha \frac{(4^{12}-2^{12})}{12}$

$EX = \frac{11}{12} \frac{4^{12}-2^{12}}{4^{11}-2^{11}} = \frac{11}{6} \frac{2^{12}-1}{2^{11}-1}$

$Answer: \frac{11}{6} \frac{2^{12}-1}{2^{11}-1}$