Exercise on finding probability density function

Question

Let $Y_1$ and $Y_2$ by independent and uniformly distributed over the interval (0, 1). Find the probability density for $U = Y_1/Y_2$:

Solution:

$F_U(u) = P(U \le u) = P(Y_1/Y_2 \le u)$. Looking at the figure:

we can see that there are two regions: $0 \le u \le 1$ and $u \gt 1$. Because $tan(\alpha) = \frac{1}{u} = \frac{y_2}{y_1}$ we see that $u = y_1$, so either can say that $F_U(u) = \frac{y_1 y_2}{2} = \frac{u y_2}{2}$, because $y_2 = 1$, then $F_U(u) = \frac{u}{2}$ for $0 \le u \le 1$. So, $f_U(u) = \frac{d[F_U(u)]}{du} = \frac{1}{2}$ on this range.

Next, we need to find $f_U(u)$ when $u \gt 1$. To do it we need to find integral:

$$F_U(u) = \int_0^1 \int_0^{y_1/u}{\frac{1}{(\theta_2 - \theta_1)^2}dy_2 dy_1} = \frac{1}{2u}$$

because $\theta_2 = 1$ and $\theta_1 = 0$. So, differentiating this, gives negative result $f_U(u) = \frac{-1}{2u^2}$. This looks explainable, because $\frac{1}{2u}$ is a hyperbola and on $u \gt 1$ it's decreasing one, so derivative is negative, but from definition of density function it cannot be negative, obviously.

Answers for this are:

$$f_U(u) = \frac{1}{2}, 0 \le u \le 1$$

and

$$f_U(u) = \frac{1}{2u^2}, u \gt 1$$

I can't understand how $f_U(u)$ turned to be positive, please, help.

Answer 1

It is simpler to do this with a single integral (using the law of total probability) rather than a double integral. For all $u>0$ we can write the distribution function as:

$$\begin{align} F_U(u) &\equiv \mathbb{P}(U \leqslant u) \\[12pt] &= \mathbb{P}(Y_1/Y_2 \leqslant u) \\[12pt] &= \mathbb{P}(Y_1 \leqslant u Y_2) \\[12pt] &= \int \limits_0^1 \mathbb{P}(Y_1 \leqslant u Y_2|Y_2 = y) \cdot \text{U}(y|0,1) \ dy \\[6pt] &= \int \limits_0^1 \mathbb{P}(Y_1 \leqslant u y) \ dy \\[6pt] &= \int \limits_0^1 \min(uy, 1) \ dy \\[6pt] &= \int \limits_0^{\min(1,1/u)} uy \ dy + \int \limits_{\min(1,1/u)}^{1} \ dy \\[12pt] &= \frac{u}{2} \min(1,1/u)^2 + \max(0,1-1/u) \\[16pt] &= \begin{cases} \frac{u}{2} & & & \text{for } 0 \leqslant u \leqslant 1, \\[6pt] \frac{u}{2} (1/u)^2 + (1-1/u) & & & \text{for } u > 1, \\[6pt] \end{cases} \\[12pt] &= \begin{cases} \frac{u}{2} & & & \text{for } 0 \leqslant u \leqslant 1, \\[6pt] 1-\frac{1}{2u} & & & \text{for } u > 1, \\[6pt] \end{cases} \\[12pt] \end{align}$$

so the density function is:

$$f_U(u) = \frac{dF_U}{du}(u) = \begin{cases} \frac{1}{2} & & & \text{for } 0 \leqslant u \leqslant 1, \\[6pt] \frac{1}{2u^2} & & & \text{for } u > 1. \\[6pt] \end{cases}$$

Answer 2

I found a solution for my question (thanks to @Ben's answer). In the figure above I need to find $P(U \le u) = P(Y_1/Y_2 \le u)$. So, I need to find area above the line $u = Y_1/Y_2$. What I found $F_U(u) = \frac{1}{2u}$ is the area below the line (I have no idea why I decided that the $P(U \le u)$ is below the line), so the final distribution function for $u \gt 1$ must be $F_U(u) = 1 - \frac{1}{2u}$, because the whole area of the square is $1$. So, I can find correct density function $f_U(u) = \frac{1}{2u^2}$.