Question 6.15: A random variable X has a mixed distribution whose discrete ......
A random variable X has a mixed distribution whose discrete and continuous parts, f_1(x) and f_2(x), are described, respectively, by
f_1(x)= \begin{cases} \frac{1}{10}, \quad x=4, \\ \frac{1}{10} , \quad x=8, \end{cases}and
f_2(x)= \begin{cases} \frac{1}{12}, \quad 0 \lt x \lt 4, \\ \frac{1}{15} , \quad 8 \lt x \leq 15. \end{cases}(i) What is the range of values for X?
(ii) Calculate the probability P(2 < X < 9).
(iii) Find the expectation and variance of X.
(i) The range for the discrete part of X is the set R_d = {4, 8}, while that for the continuous part, as can be seen from the formula for f_2 above, is the union of the intervals (0, 4) and (8, 15]. The range of X is thus the union R_d \cup R_c, which is the set R_X = (0, 4]∪[8, 15].
(ii) We use formula (6.17),
\sum\limits_{\mathrm{discrete \,part}}f_{1}(x)+\int_{\mathrm{continuous\, part}}f_{2}(x)\mathrm{d}x=1.\qquad \qquad (6.17)separating the discrete from the continuous part of X, which gives
P(2\lt X\lt 9)=P(2\lt X\lt 4)+P(X=4)+P(4\lt X\lt 8)+ P(X = 8) + P(8 \lt X \lt 9)\\ =\int_{2}^{4}f_{2}(x)\mathrm{d}x+f_{1}(4)+\int_{4}^{8}f_{2}(x)\mathrm{d}x+f_{1}(8)+\int_{8}^{9}f_{2}(x)\mathrm{d}x\\ =\int_{2}^{4}{\frac{1}{12}}\mathrm{d}x+{\frac{1}{10}}+0+{\frac{1}{10}}+\int_{8}^{9}{\frac{1}{15}}\mathrm{d}x\\ ={\frac{1}{12}}\cdot(4-2)+{\frac{1}{10}}+0+{\frac{1}{10}}+{\frac{1}{15}}\cdot(9-8)={\frac{13}{30}}.(iii) In order to find the expectation of X, we have to consider again separately the discrete and the continuous parts of the distribution. More specifically, we have
E(X)=\sum\limits_{\mathrm{discrete\, part}}x f_{1}(x)+\int_{\mathrm{continuous \,part}}x f_{2}(x)\mathrm{d}x\\ =4\cdot{\frac{1}{10}}+8\cdot{\frac{1}{10}}+\int_{0}^{4}{\frac{x}{12}}\,\mathrm{d}x+\int_{8}^{15}{\frac{x}{15}}\,\mathrm{d}x\\ ={\frac{2}{5}}+{\frac{4}{5}}+{\frac{1}{12}}\cdot\left[{\frac{x^{2}}{2}}\right]_{0}^{4}+{\frac{1}{15}}\cdot\left[{\frac{x^{2}}{2}}\right]_{8}^{15}\\ ={\frac{2}{5}}+{\frac{4}{5}}+{\frac{8}{12}}+{\frac{15^{2}-8^{2}}{2\cdot15}}={\frac{217}{30}}\cong7.23.For the variance of X, in a similar way, we first find E(X²) as
E(X^{2})=\sum\limits_{\mathrm{discrete\, part}}x^{2}f_{1}(x)+\int_{\mathrm{continuous\, part}}x^{2}f_{2}(x)\mathrm{d}x\\ =4^{2}\cdot{\frac{1}{10}}+8^{2}\cdot{\frac{1}{10}}+\int_{0}^{4}{\frac{x^{2}}{12}}\mathrm{d}x+\int_{8}^{15}{\frac{x^{2}}{15}}\mathrm{d}x\\ =\frac{8}{5}+\frac{32}{5}+\frac{1}{12}\cdot\left[\frac{x^{3}}{3}\right]_{0}^{4}+\frac{1}{15}\cdot\left[\frac{x^{3}}{3}\right]_{8}^{15}=\frac{367}{5}=73.4.We therefore find the variance of X to be
\mathrm{Var}(X) = E(X^2)−[E(X)]^2 = 73.4 − (7.23)^2 = 21.08.