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I'm going wrong with the calculation of CDF for the ratio $\frac{Y}X$ where $X$ and $Y$ are follow uniform distribution in range (0,1).

Let $ T = \frac{Y}X $

CDF of T is $F_T(t) = P(\frac{Y}X <t)$

To calculate $F_T(t)$ condition on $X$ and then apply Law of total probability

$P(\frac{Y}X <t) = \int P(Y<tx|X=x)P(X=x)dx$ and $P(X=x)=1$ since $X$ and $Y$ are independent

$P(\frac{Y}X <t) = \int P(Y<tx)dx$

Range of $tx$ can be either $0<tx<1$ or $tx>1$
Then calculate $F_T(t)$ for above two cases of $tx$

For $0<tx<1$ integral becomes $P(\frac{Y}X <t) = \int_0^1 P(Y<tx)dx = \int_0^1 txdx = \frac{t}2$

For $tx>1 P(Y<tx)$ becomes 1,so $P(\frac{Y}X <t) = \int_0^1 1dx = \int_0^1 txdx = 1$
I know that the problem is with the limits of integral, this post answers the same question, but doesn't explain how those limits were arrived at.

Effective_cellist
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1 Answers1

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As you say, $\mathsf P(Y<tx) = \begin{cases} tx &:& 0\leq tx<1\\ 1 &:& tx\geq 1 \\ 0 &:& tx<0\end{cases}$

However, consider when does this happen if you are integrating with respect to $x\in [0;1]$ ?

$\begin{align}F_T(t)~&=~\mathsf P(Y<tX)\\ &=~ \int_0^1 \mathsf P(Y<tx\mid X=x)f_X(x)\operatorname d x \\ &=~ \int_0^1 \mathsf P(Y<tx)\operatorname d x\\ &=~\int_0^1 \min\{1,tx\}\operatorname d x \\ &=~\begin{cases}\int_0^{1/t}tx\operatorname d x+\int_{1/t}^1 1\operatorname d x &:& 0\leq 1/t< 1\\ \int_0^{1}tx\operatorname d x &:& 1\leq 1/t\end{cases} \\ &~~\vdots\end{align}$

Graham Kemp
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