Question 7.9: If X is the amount of money spent on food and other expenses......
If X is the amount of money spent on food and other expenses during a day (in €) and Y is the daily allowance of a businesswoman, the joint density of these two variables is given by
f_{XY} (x, y) =\left\{\begin{matrix} \quad c\left(\frac{100-x}{x} \right) \quad\quad if 10 \leq x \leq 100, 40 \leq y \leq 100\\ 0 \quad\quad \quad\quad elsewhere.\end{matrix} \right.
(a) Choose c such that f_{XY} (x, y) is a probability density function.
(b) Find the marginal distribution of X.
(c) Calculate the probability that more than €75 are spent.
(d) Determine the conditional distribution of Y given X.
(a) The constant c must satisfy
\int_{40}^{100}{\int_{10}^{100}{c\left(\frac{100-x}{x} \right)}dx dy } = \int_{40}^{100}{\int_{10}^{100}{\frac{100c}{x}-c } dx dy} \overset{!}{=} 1
and therefore
\int_{40}^{100}{\left[100c \ln \left(x\right) -cx\right] ^{100}_{10} dy} = \int_{40}^{100}{100c \ln 100 -100c-100c\ln \left(10\right)+10c dy}
which is
\int_{40}^{100}{100c \left(\ln \frac{100}{10}-1+\frac{1}{10}\right) }dy = \left[100cy \left(\ln10 – {9}/{10}\right)\right] ^{100}_{40}
= 600c(10 ln 10 − 9) → c ≈ 0.00012.
(b) The marginal distribution is
f_{X}\left(x\right) =\int_{40}^{100}{c \left(\frac{100-x}{x} \right) dy} =\left[c \left(\frac{100-x}{x} \right)y\right] ^{100}_{40}
=100 c \left(\frac{100-x}{x} \right)-40c \left(\frac{100-x}{x} \right)\approx 0.00713 \left(\frac{100-x}{x} \right)
for 10 ≤ x ≤ 100.
(c) To determine P(X > 75), we need the cumulative marginal distribution of X:
F_{X}\left(x\right) =\int_{-\infty }^{x}{f_{X}\left(t\right)dt} =\int_{10}^{x}{0.00713 \left(\frac{100-t}{t} \right)dt}
=\int_{10}^{x}{\frac{0.00713}{t} – 0.00713 dt} =\left[0.713\ln \left(t\right) -0.00713\right] ^{x}_{10}
= 0.713 ln(x) − 0.00713x − 0.00713 ln(10) + 0.00713 · 10.
Now we can calculate
P(X > 75) = 1 − P(X ≤ 75) = 1 − F_{X}(75) = 1 − 0.973 ≈ 2.7 %.
(d) The conditional distribution is
f_{Y\mid X}\left(x,y\right)=\frac{f\left(x,y\right)}{f\left(x\right)} =\frac{c\left(\frac{100-x}{x} \right)}{60c \left(\frac{100-x}{x} \right)} =\frac{1}{60} .