STEM Problem Detail

Problem #85

Statistics Hard

Problem Statement

Prompt

Provide me the step-by-step solution to Question 17.

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The joint probability density function of $$(X, Y)$$ is
$$f_{X,Y}(x, y) = x + y, \quad 0 < x < 1, 0 < y < 1,$$and $$f_{X,Y}(x, y) = 0$$ otherwise. Let $$U = XY$$.

For $$0 < u < 1$$,
$$F_U(u) = P(XY \le u) = \int_0^1 \int_0^{\min(1, \frac{u}{x})} (x + y) dy dx.$$Split the integral according to whether $$\frac{u}{x} \ge 1$$ or $$\frac{u}{x} < 1$$.
When $$0 < x \le u$$, $$\frac{u}{x} \ge 1$$, so $$y$$ varies from $$0$$ to $$1$$.
When $$u < x < 1$$, $$\frac{u}{x} < 1$$, so $$y$$ varies from $$0$$ to $$\frac{u}{x}$$.

Thus,
$$F_U(u) = \int_0^u \int_0^1 (x + y) dy dx + \int_u^1 \int_0^{u/x} (x + y) dy dx.$$
Evaluating,
$$\int_0^u \int_0^1 (x + y) dy dx = \int_0^u (x + \frac{1}{2}) dx = \frac{u^2}{2} + \frac{u}{2},$$and
$$\int_u^1 \int_0^{u/x} (x + y) dy dx = \int_u^1 (x \frac{u}{x} + \frac{1}{2} \frac{u^2}{x^2}) dx = \int_u^1 (u + \frac{u^2}{2x^2}) dx = u(1 - u) + \frac{u^2}{2} (\frac{1}{u} - 1) = u - u^2 + \frac{u}{2} - \frac{u^2}{2}.$$Hence,
$$F_U(u) = (\frac{u^2}{2} + \frac{u}{2}) + (u - u^2 + \frac{u}{2} - \frac{u^2}{2}) = u^2 + u - \frac{3}{2}u^2 = 2u - u^2, \quad 0 < u < 1.$$
Differentiating,
$$f_U(u) = \frac{d}{du}F_U(u) = 2 - 2u = 2(1 - u), \quad 0 < u < 1,$$and $$f_U(u) = 0$$ otherwise.

$$f_U(u) = 2(1 - u), \quad 0 < u < 1.$$

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