Intuitive Derivation

Clearly, if $x \geq 1$, then $P(X < x | Z) = 1$ no matter what the observed value of $Z$ is.

If $0 \leq x < 1$ and suppose we have observed $Z = z$. Intuitively, it holds that (this actually has a theoretical support too: see Problem 33.16 in Probability and Measure (3rd ed.) by Patrick Billingsley): \begin{align*} P(X < x | Z = z) &= \lim_{h \downarrow 0} P(X < x | z - h < Z \leq z + h) \\ &= \lim_{h \downarrow 0} P(X < x | (z - h)X < Y \leq (z + h)X) \\ &= \lim_{h \downarrow 0} \frac{P(X < x, (z - h)X < Y \leq (z + h)X)}{P((z - h)X < Y \leq (z + h)X)}. \tag{$\star$}\label{star} \end{align*}

Case 1:

If $0 < z < 1$, then for sufficiently small $h$, $0 < z - h < z + h < 1$, thus the probability $P((z - h)X < Y \leq (z + h)X)$ is the area of the region (draw a picture) \begin{align*} \{(u, v): 0 < u < 1, (z - h)u < v \leq (z + h)u\}, \end{align*} which is $\frac{1}{2} \times [(z + h) - (z - h)] \times 1 = h$.

Similarly, the probability $P(X < x, (z - h)X < Y \leq (z + h)X)$ is the area of the region \begin{align*} \{(u, v): 0 < u < x, (z - h)u < v \leq (z + h)u\}, \end{align*} which is $\frac{1}{2} \times [(z + h)x - (z - h)x] \times x = hx^2$. It thus follows by $\eqref{star}$ that \begin{align*} P(X < x | Z = z) = \lim_{h \to 0}\frac{hx^2}{h} = x^2. \end{align*}

Case 2:

If $z > 1$, this case is slightly more involved than Case 1, as $\eqref{star}$ depends on the order of $x$ and $z^{-1}$. While the probability of $P((z - h)X < Y < (z + h)X)$ is always (again, draw the picture) \begin{align*} & \operatorname{Area}(S_1) := \operatorname{Area}(\{(u, v): (z + h)^{-1}v < u < (z - h)^{-1}v, 0 < v < 1\}) \\ =& \frac{1}{2} \times \left(\frac{1}{z - h} - \frac{1}{z + h}\right) \times 1 = \frac{h}{z^2 - h^2}, \end{align*} the shape of the region $\{(u, v): 0 < u < x\} \cap S_1$ is different depending on whether $x \leq z^{-1}$ or $x > z^{-1}$. One can verify that \begin{align*} & P(X < x, (z - h)X < Y \leq (z + h)X) \\ =& \begin{cases} \frac{1}{2} \times [(z + h)x - (z - h)x] \times x = hx^2 & x < z^{-1}, \\ \frac{1}{2} \times \left(\frac{1}{z - h} - \frac{1}{z + h}\right) \times 1 = \frac{h}{z^2 - h^2} & x \geq z^{-1}. \end{cases} \end{align*} It thus follows by $\eqref{star}$ that \begin{align*} P(X < x | Z = z) = \begin{cases} \lim_{h \downarrow 0}\frac{hx^2}{\frac{h}{z^2 - h^2}} = x^2z^2 & x < z^{-1}, \\ 1 & x \geq z^{-1}. \end{cases} = \min(x^2z^2, 1). \end{align*}

In summary, Case 1 and Case 2 show that \begin{align*} P(X < x | Z) = \min(x^2, 1)I_{(0, 1]}(Z) + \min(x^2Z^2, 1)I_{[1, \infty)}(Z). \tag{$\dagger$}\label{dagger} \end{align*}

Rigorous Proof

Now we verify $\eqref{dagger}$ by checking it satisfies the two defining properties of conditional probability:

1. $P(X < x |Z)$ is measurable $\sigma(Z)$.
2. For any $G \in \sigma(Z)$, $\int_G P(X < x | Z)dP = P(G \cap \{X < x\})$.

Bullet point 1 is obvious in view of $\eqref{dagger}$ is a function of $Z$. To verify bullet point 2, we need to show for any $z \in (0, +\infty)$, \begin{align*} \int\limits_{Z \leq z}\left[\min(x^2, 1)I_{(0, 1]}(Z) + \min(x^2Z^2, 1)I_{[1, \infty)}(Z)\right]dP = P(X \leq x, Z \leq z). \end{align*} Or equivalently, \begin{align*} \int_0^z\left[\min(x^2, 1)I_{(0, 1]}(t) + \min(x^2t^2, 1)I_{[1, \infty)}(t)\right]f_Z(t)dt = P(X \leq x, Z \leq z), \tag{0}\label{0} \end{align*} where $f_Z$ denotes the density of $Z = \frac{Y}{X}$. By the independence of $X$ and $Y$, the distribution function of $Z$ is \begin{align*} F_Z(z) = P(Y \leq zX) = \int_0^1P(Y \leq zx)dx. \tag{1}\label{1} \end{align*}

If $z > 1$, the integral in $\eqref{1}$ is \begin{align*} & \int_0^{z^{-1}}P(Y \leq zx)dx + \int_{z^{-1}}^1P(Y \leq zx)dx \\ =& \int_0^{z^{-1}}zxdx + \int_{z^{-1}}^11dx = \frac{1}{2z} + 1 - \frac{1}{z} = 1 - \frac{1}{2z}. \tag{2}\label{2} \end{align*}

If $z \leq 1$, the integral in $\eqref{1}$ is \begin{align*} \int_0^1 zx dx = \frac{1}{2}z. \tag{3}\label{3} \end{align*} $\eqref{2}$ and $\eqref{3}$ thus imply the density of $Z$ is given by \begin{align*} f_Z(z) = \begin{cases} \frac{1}{2} & z \leq 1, \\ \frac{1}{2}z^{-2} & z > 1. \end{cases} \tag{4}\label{4} \end{align*}

When $z \leq 1$, it follows by $\eqref{4}$ that the left hand side of $\eqref{0}$ is \begin{align*} \int_0^z \min(x^2, 1)\frac{1}{2}dt = \frac{1}{2}\min(x^2, 1)z, \end{align*} while the right hand side of $\eqref{0}$ is \begin{align*} P(X \leq x, Y \leq Xz) = \int_0^1 P(t \leq x, Y \leq tz)dt = \frac{1}{2}z\min(x^2, 1). \end{align*} Hence $\eqref{0}$ holds.

When $z > 1$, it follows by $\eqref{4}$ that the left hand side of $\eqref{0}$ is \begin{align*} & \int_0^1 \min(x^2, 1)\frac{1}{2}dt + \int_1^z \min(x^2t^2, 1) \frac{1}{2}t^{-2}dt \\ =& \frac{1}{2}\min(x^2, 1) + \frac{1}{2}\int_1^z \min(x^2, t^{-2})dt \\ =& \begin{cases} \frac{1}{2}zx^2 & x \leq z^{-1}, \\ x - \frac{1}{2z} & z^{-1} < x < 1, \\ 1 - \frac{1}{2z} & x \geq 1, \end{cases} \end{align*} while the right hand side of $\eqref{0}$ is \begin{align*} & P(X \leq x, Y \leq Xz) = \int_0^1 P(t \leq x, Y \leq tz)dt \\ =& \begin{cases} \frac{1}{2}zx^2 & x \leq z^{-1}, \\ x - \frac{1}{2z} & z^{-1} < x < 1, \\ 1 - \frac{1}{2z} & x \geq 1. \end{cases} \end{align*} Hence $\eqref{0}$ holds. This completes the proof.

Intuitive Derivation

Clearly, if $x \geq 1$, then $P(X < x | Z) = 1$ no matter what the observed value of $Z$ is.

If $0 \leq x < 1$ and suppose we have observed $Z = z$. Intuitively, it holds that (this actually has a theoretical support too: see Problem 33.16 in Probability and Measure (3rd ed.) by Patrick Billingsley): \begin{align*} P(X < x | Z = z) &= \lim_{h \downarrow 0} P(X < x | z - h < Z \leq z + h) \\ &= \lim_{h \downarrow 0} P(X < x | (z - h)X < Y \leq (z + h)X) \\ &= \lim_{h \downarrow 0} \frac{P(X < x, (z - h)X < Y \leq (z + h)X)}{P((z - h)X < Y \leq (z + h)X)}. \tag{$\star$}\label{star} \end{align*}

Case 1:

If $0 < z < 1$, then for sufficiently small $h$, $0 < z - h < z + h < 1$, thus the probability $P((z - h)X < Y \leq (z + h)X)$ is the area of the region (draw a picture) \begin{align*} \{(u, v): 0 < u < 1, (z - h)u < v \leq (z + h)u\}, \end{align*} which is $\frac{1}{2} \times [(z + h) - (z - h)] \times 1 = h$.

Similarly, the probability $P(X < x, (z - h)X < Y \leq (z + h)X)$ is the area of the region \begin{align*} \{(u, v): 0 < u < x, (z - h)u < v \leq (z + h)u\}, \end{align*} which is $\frac{1}{2} \times [(z + h)x - (z - h)x] \times x = hx^2$. It thus follows by $\eqref{star}$ that \begin{align*} P(X < x | Z = z) = \lim_{h \to 0}\frac{hx^2}{h} = x^2. \end{align*}

Case 2:

If $z > 1$, this case is slightly more involved than Case 1, as $\eqref{star}$ depends on the order of $x$ and $z^{-1}$. While the probability of $P((z - h)X < Y < (z + h)X)$ is always (again, draw the picture) \begin{align*} & \operatorname{Area}(S_1) := \operatorname{Area}(\{(u, v): (z + h)^{-1}v < u < (z - h)^{-1}v, 0 < v < 1\}) \\ =& \frac{1}{2} \times \left(\frac{1}{z - h} - \frac{1}{z + h}\right) \times 1 = \frac{h}{z^2 - h^2}, \end{align*} the shape of the region $\{(u, v): 0 < u < x\} \cap S_1$ is different depending on whether $x \leq z^{-1}$ or $x > z^{-1}$. One can verify that \begin{align*} & P(X < x, (z - h)X < Y \leq (z + h)X) \\ =& \begin{cases} \frac{1}{2} \times [(z + h)x - (z - h)x] \times x = hx^2 & x < z^{-1}, \\ \frac{1}{2} \times \left(\frac{1}{z - h} - \frac{1}{z + h}\right) \times 1 = \frac{h}{z^2 - h^2} & x \geq z^{-1}. \end{cases} \end{align*} It thus follows by $\eqref{star}$ that \begin{align*} P(X < x | Z = z) = \begin{cases} \lim_{h \downarrow 0}\frac{hx^2}{\frac{h}{z^2 - h^2}} = x^2z^2 & x < z^{-1} \\ 1 & x \geq z^{-1} \end{cases} = \min(x^2z^2, 1). \end{align*}

In summary, Case 1 and Case 2 show that \begin{align*} P(X < x | Z) = \min(x^2, 1)I_{(0, 1]}(Z) + \min(x^2Z^2, 1)I_{[1, \infty)}(Z). \tag{$\dagger$}\label{dagger} \end{align*}

Rigorous Proof

Now we verify $\eqref{dagger}$ by checking it satisfies the two defining properties of conditional probability:

1. $P(X < x |Z)$ is measurable $\sigma(Z)$.
2. For any $G \in \sigma(Z)$, $\int_G P(X < x | Z)dP = P(G \cap \{X < x\})$.

Bullet point 1 is obvious in view of $\eqref{dagger}$ is a function of $Z$. To verify bullet point 2, we need to show for any $z \in (0, +\infty)$, \begin{align*} \int\limits_{Z \leq z}\left[\min(x^2, 1)I_{(0, 1]}(Z) + \min(x^2Z^2, 1)I_{[1, \infty)}(Z)\right]dP = P(X \leq x, Z \leq z). \end{align*} Or equivalently, \begin{align*} \int_0^z\left[\min(x^2, 1)I_{(0, 1]}(t) + \min(x^2t^2, 1)I_{[1, \infty)}(t)\right]f_Z(t)dt = P(X \leq x, Z \leq z), \tag{0}\label{0} \end{align*} where $f_Z$ denotes the density of $Z = \frac{Y}{X}$. By the independence of $X$ and $Y$, the distribution function of $Z$ is \begin{align*} F_Z(z) = P(Y \leq zX) = \int_0^1P(Y \leq zx)dx. \tag{1}\label{1} \end{align*}

If $z > 1$, the integral in $\eqref{1}$ is \begin{align*} & \int_0^{z^{-1}}P(Y \leq zx)dx + \int_{z^{-1}}^1P(Y \leq zx)dx \\ =& \int_0^{z^{-1}}zxdx + \int_{z^{-1}}^11dx = \frac{1}{2z} + 1 - \frac{1}{z} = 1 - \frac{1}{2z}. \tag{2}\label{2} \end{align*}

If $z \leq 1$, the integral in $\eqref{1}$ is \begin{align*} \int_0^1 zx dx = \frac{1}{2}z. \tag{3}\label{3} \end{align*} $\eqref{2}$ and $\eqref{3}$ thus imply the density of $Z$ is given by \begin{align*} f_Z(z) = \begin{cases} \frac{1}{2} & z \leq 1, \\ \frac{1}{2}z^{-2} & z > 1. \end{cases} \tag{4}\label{4} \end{align*}

When $z \leq 1$, it follows by $\eqref{4}$ that the left hand side of $\eqref{0}$ is \begin{align*} \int_0^z \min(x^2, 1)\frac{1}{2}dt = \frac{1}{2}\min(x^2, 1)z, \end{align*} while the right hand side of $\eqref{0}$ is \begin{align*} P(X \leq x, Y \leq Xz) = \int_0^1 P(t \leq x, Y \leq tz)dt = \frac{1}{2}z\min(x^2, 1). \end{align*} Hence $\eqref{0}$ holds.

When $z > 1$, it follows by $\eqref{4}$ that the left hand side of $\eqref{0}$ is \begin{align*} & \int_0^1 \min(x^2, 1)\frac{1}{2}dt + \int_1^z \min(x^2t^2, 1) \frac{1}{2}t^{-2}dt \\ =& \frac{1}{2}\min(x^2, 1) + \frac{1}{2}\int_1^z \min(x^2, t^{-2})dt \\ =& \begin{cases} \frac{1}{2}zx^2 & x \leq z^{-1}, \\ x - \frac{1}{2z} & z^{-1} < x < 1, \\ 1 - \frac{1}{2z} & x \geq 1, \end{cases} \end{align*} while the right hand side of $\eqref{0}$ is \begin{align*} & P(X \leq x, Y \leq Xz) = \int_0^1 P(t \leq x, Y \leq tz)dt \\ =& \begin{cases} \frac{1}{2}zx^2 & x \leq z^{-1}, \\ x - \frac{1}{2z} & z^{-1} < x < 1, \\ 1 - \frac{1}{2z} & x \geq 1. \end{cases} \end{align*} Hence $\eqref{0}$ holds. This completes the proof.

Intuitive Derivation

Clearly, if $x \geq 1$, then $P(X < x | Z) = 1$ no matter what the observed value of $Z$ is.

If $0 \leq x < 1$ and suppose we have observed $Z = z$. Let's consider the joint distribution of $(\xi, \eta)$, where $\xi = X, \eta = Y - zX$. There are two sub-cases:

• $z \leq 1$. In this case, the support of $(\xi, \eta)$ is the parallelogram $$S = \{(u, v): 0 < u < 1, -z < v < 1 - z\},$$ and the joint density of $(\xi, \eta)$ is given by (use the independence of $X$ and $Y$ here): \begin{align*} g(u, v) = \begin{cases} 1 & (u, v) \in S, \\ 0 & (u, v) \notin S. \end{cases} \end{align*} Consequently, the conditional density of $\xi$ given $\eta = 0$ is \begin{align*} g_{\xi|\eta}(u|\eta = 0) = \frac{g(u, 0)}{g_\eta(0)} = 1, \; 0 < u < 1. \end{align*} The conditional probability of interest is then \begin{align*} P(X < x | Y = zX) = P(\xi < x | \eta = 0) = \int_0^x 1 du = x. \end{align*}

Rigorous Proof

For any $z \in (0, +\infty)$, we need to show \begin{align*} \int\limits_{Z \leq z}\left[\min(x^2, 1)I_{(0, 1]}(Z) + \min(x^2Z^2, 1)I_{[1, \infty)}(Z)\right]dP = P(X \leq x, Z \leq z). \end{align*} Or equivalently, \begin{align*} \int_0^z\left[\min(x^2, 1)I_{(0, 1]}(t) + \min(x^2t^2, 1)I_{[1, \infty)}(t)\right]f_Z(t)dt = P(X \leq x, Z \leq z), \tag{0}\label{0} \end{align*} where $f_Z$ denotes the density of $Z = \frac{Y}{X}$. By the independence of $X$ and $Y$, the distribution function of $Z$ is \begin{align*} F_Z(z) = P(Y \leq zX) = \int_0^1P(Y \leq zx)dx. \tag{1}\label{1} \end{align*}

If $z > 1$, the integral in $\eqref{1}$ is \begin{align*} & \int_0^{z^{-1}}P(Y \leq zx)dx + \int_{z^{-1}}^1P(Y \leq zx)dx \\ =& \int_0^{z^{-1}}zxdx + \int_{z^{-1}}^11dx = \frac{1}{2z} + 1 - \frac{1}{z} = 1 - \frac{1}{2z}. \tag{2}\label{2} \end{align*}

If $z \leq 1$, the integral in $\eqref{1}$ is \begin{align*} \int_0^1 zx dx = \frac{1}{2}z. \tag{3}\label{3} \end{align*} $\eqref{2}$ and $\eqref{3}$ thus imply the density of $Z$ is given by \begin{align*} f_Z(z) = \begin{cases} \frac{1}{2} & z \leq 1, \\ \frac{1}{2}z^{-2} & z > 1. \end{cases} \tag{4}\label{4} \end{align*}

When $z \leq 1$, it follows by $\eqref{4}$ that the left hand side of $\eqref{0}$ is \begin{align*} \int_0^z \min(x^2, 1)\frac{1}{2}dt = \frac{1}{2}\min(x^2, 1)z, \end{align*} while the right hand side of $\eqref{0}$ is \begin{align*} P(X \leq x, Y \leq Xz) = \int_0^1 P(t \leq x, Y \leq tz)dt = \frac{1}{2}z\min(x^2, 1). \end{align*} Hence $\eqref{0}$ holds.

When $z > 1$, it follows by $\eqref{4}$ that the left hand side of $\eqref{0}$ is \begin{align*} & \int_0^1 \min(x^2, 1)\frac{1}{2}dt + \int_1^z \min(x^2t^2, 1) \frac{1}{2}t^{-2}dt \\ =& \frac{1}{2}\min(x^2, 1) + \frac{1}{2}\int_1^z \min(x^2, t^{-2})dt \\ =& \begin{cases} \frac{1}{2}zx^2 & x \leq z^{-1}, \\ x - \frac{1}{2z} & z^{-1} < x < 1, \\ 1 - \frac{1}{2z} & x \geq 1, \end{cases} \end{align*} while the right hand side of $\eqref{0}$ is \begin{align*} & P(X \leq x, Y \leq Xz) = \int_0^1 P(t \leq x, Y \leq tz)dt \\ =& \begin{cases} \frac{1}{2}zx^2 & x \leq z^{-1}, \\ x - \frac{1}{2z} & z^{-1} < x < 1, \\ 1 - \frac{1}{2z} & x \geq 1. \end{cases} \end{align*} Hence $\eqref{0}$ holds. This completes the proof.

Intuitive Derivation

Clearly, if $x \geq 1$, then $P(X < x | Z) = 1$ no matter what the observed value of $Z$ is.

If $0 \leq x < 1$ and suppose we have observed $Z = z$. Intuitively, it holds that (this actually has a theoretical support too: see Problem 33.16 in Probability and Measure (3rd ed.) by Patrick Billingsley): \begin{align*} P(X < x | Z = z) &= \lim_{h \downarrow 0} P(X < x | z - h < Z \leq z + h) \\ &= \lim_{h \downarrow 0} P(X < x | (z - h)X < Y \leq (z + h)X) \\ &= \lim_{h \downarrow 0} \frac{P(X < x, (z - h)X < Y \leq (z + h)X)}{P((z - h)X < Y \leq (z + h)X)}. \tag{$\star$}\label{star} \end{align*}

Case 1:

If $0 < z < 1$, then for sufficiently small $h$, $0 < z - h < z + h < 1$, thus the probability $P((z - h)X < Y \leq (z + h)X)$ is the area of the region (draw a picture) \begin{align*} \{(u, v): 0 < u < 1, (z - h)u < v \leq (z + h)u\}, \end{align*} which is $\frac{1}{2} \times [(z + h) - (z - h)] \times 1 = h$.

Similarly, the probability $P(X < x, (z - h)X < Y \leq (z + h)X)$ is the area of the region \begin{align*} \{(u, v): 0 < u < x, (z - h)u < v \leq (z + h)u\}, \end{align*} which is $\frac{1}{2} \times [(z + h)x - (z - h)x] \times x = hx^2$. It thus follows by $\eqref{star}$ that \begin{align*} P(X < x | Z = z) = \lim_{h \to 0}\frac{hx^2}{h} = x^2. \end{align*}

Case 2:

If $z > 1$, this case is slightly more involved than Case 1, as $\eqref{star}$ depends on the order of $x$ and $z^{-1}$. While the probability of $P((z - h)X < Y < (z + h)X)$ is always (again, draw the picture) \begin{align*} & \operatorname{Area}(S_1) := \operatorname{Area}(\{(u, v): (z + h)^{-1}v < u < (z - h)^{-1}v, 0 < v < 1\}) \\ =& \frac{1}{2} \times \left(\frac{1}{z - h} - \frac{1}{z + h}\right) \times 1 = \frac{h}{z^2 - h^2}, \end{align*} the shape of the region $\{(u, v): 0 < u < x\} \cap S_1$ is different depending on whether $x \leq z^{-1}$ or $x > z^{-1}$. One can verify that \begin{align*} & P(X < x, (z - h)X < Y \leq (z + h)X) \\ =& \begin{cases} \frac{1}{2} \times [(z + h)x - (z - h)x] \times x = hx^2 & x < z^{-1}, \\ \frac{1}{2} \times \left(\frac{1}{z - h} - \frac{1}{z + h}\right) \times 1 = \frac{h}{z^2 - h^2} & x \geq z^{-1}. \end{cases} \end{align*} It thus follows by $\eqref{star}$ that \begin{align*} P(X < x | Z = z) = \begin{cases} \lim_{h \downarrow 0}\frac{hx^2}{\frac{h}{z^2 - h^2}} = x^2z^2 & x < z^{-1}, \\ 1 & x \geq z^{-1}. \end{cases} = \min(x^2z^2, 1). \end{align*}

In summary, Case 1 and Case 2 show that \begin{align*} P(X < x | Z) = \min(x^2, 1)I_{(0, 1]}(Z) + \min(x^2Z^2, 1)I_{[1, \infty)}(Z). \tag{$\dagger$}\label{dagger} \end{align*}

Rigorous Proof

Now we verify $\eqref{dagger}$ by checking it satisfies the two defining properties of conditional probability:

1. $P(X < x |Z)$ is measurable $\sigma(Z)$.
2. For any $G \in \sigma(Z)$, $\int_G P(X < x | Z)dP = P(G \cap \{X < x\})$.

Bullet point 1 is obvious in view of $\eqref{dagger}$ is a function of $Z$. To verify bullet point 2, we need to show for any $z \in (0, +\infty)$, \begin{align*} \int\limits_{Z \leq z}\left[\min(x^2, 1)I_{(0, 1]}(Z) + \min(x^2Z^2, 1)I_{[1, \infty)}(Z)\right]dP = P(X \leq x, Z \leq z). \end{align*} Or equivalently, \begin{align*} \int_0^z\left[\min(x^2, 1)I_{(0, 1]}(t) + \min(x^2t^2, 1)I_{[1, \infty)}(t)\right]f_Z(t)dt = P(X \leq x, Z \leq z), \tag{0}\label{0} \end{align*} where $f_Z$ denotes the density of $Z = \frac{Y}{X}$. By the independence of $X$ and $Y$, the distribution function of $Z$ is \begin{align*} F_Z(z) = P(Y \leq zX) = \int_0^1P(Y \leq zx)dx. \tag{1}\label{1} \end{align*}

If $z > 1$, the integral in $\eqref{1}$ is \begin{align*} & \int_0^{z^{-1}}P(Y \leq zx)dx + \int_{z^{-1}}^1P(Y \leq zx)dx \\ =& \int_0^{z^{-1}}zxdx + \int_{z^{-1}}^11dx = \frac{1}{2z} + 1 - \frac{1}{z} = 1 - \frac{1}{2z}. \tag{2}\label{2} \end{align*}

If $z \leq 1$, the integral in $\eqref{1}$ is \begin{align*} \int_0^1 zx dx = \frac{1}{2}z. \tag{3}\label{3} \end{align*} $\eqref{2}$ and $\eqref{3}$ thus imply the density of $Z$ is given by \begin{align*} f_Z(z) = \begin{cases} \frac{1}{2} & z \leq 1, \\ \frac{1}{2}z^{-2} & z > 1. \end{cases} \tag{4}\label{4} \end{align*}

When $z \leq 1$, it follows by $\eqref{4}$ that the left hand side of $\eqref{0}$ is \begin{align*} \int_0^z \min(x^2, 1)\frac{1}{2}dt = \frac{1}{2}\min(x^2, 1)z, \end{align*} while the right hand side of $\eqref{0}$ is \begin{align*} P(X \leq x, Y \leq Xz) = \int_0^1 P(t \leq x, Y \leq tz)dt = \frac{1}{2}z\min(x^2, 1). \end{align*} Hence $\eqref{0}$ holds.

When $z > 1$, it follows by $\eqref{4}$ that the left hand side of $\eqref{0}$ is \begin{align*} & \int_0^1 \min(x^2, 1)\frac{1}{2}dt + \int_1^z \min(x^2t^2, 1) \frac{1}{2}t^{-2}dt \\ =& \frac{1}{2}\min(x^2, 1) + \frac{1}{2}\int_1^z \min(x^2, t^{-2})dt \\ =& \begin{cases} \frac{1}{2}zx^2 & x \leq z^{-1}, \\ x - \frac{1}{2z} & z^{-1} < x < 1, \\ 1 - \frac{1}{2z} & x \geq 1, \end{cases} \end{align*} while the right hand side of $\eqref{0}$ is \begin{align*} & P(X \leq x, Y \leq Xz) = \int_0^1 P(t \leq x, Y \leq tz)dt \\ =& \begin{cases} \frac{1}{2}zx^2 & x \leq z^{-1}, \\ x - \frac{1}{2z} & z^{-1} < x < 1, \\ 1 - \frac{1}{2z} & x \geq 1. \end{cases} \end{align*} Hence $\eqref{0}$ holds. This completes the proof.