Random Ratio

We have 2 random variables (RVs) $p$ and $q$ here. Let's try to fix one for simplicity, and then use that result to develop a general solution where both are RVs.

Step 1: Fixing q

Say we fix $q$ at some value between 0 and 1. Now, the given problem reduces to:

"q is a fixed value between 0 and 1. p ~ Uniform(0,1). What is the probability that 1 < p/q < 2?"

This question is fairly straightforward.

Step 2: Rearranging the Inequality

Let's rearrange the inequality:

$$1<\frac{p}{q}<2$$ $$q<p<2q$$

Also, note that since p is restricted in (0,1), we must say:

$$\max (0,q)<p<\min (2q,1)$$

But since q is also restricted in (0,1), we have

$$\max (0,q)=q$$

Thus, the final expression is:

$$q<p<\min (2q,1)$$

Step 3: Computing the Probability

Now, let's compute the probability that p will lie within this range.

As

$$p\sim \text{Uniform}(0,1)$$

the required probability:

$$P=\frac{\min (1,2q)-q}{1}=\min (2q,1)-q$$

(Note: The probability is simply the length of the required range divided by the length of the total range. This is also visible by intuition)

Step 4: Considering q as a Random Variable

Now, note that this is the case when q is fixed. But q itself will vary from 0 to 1.

We can write the above probability as "conditioned on" a particular q:

$$P(E|q)=\min (2q,1)-q$$

(Where E is our required event of

$$1<\frac{p}{q}<2$$

)

Step 5: Using the Law of Total Probability

Using fundamental laws, we have:

$$P(E)=\sum P(E|q)P(q)$$

What is P(q)? Essentially the probability of choosing a particular value from 0 to 1, an infinite set! Clearly very small! Let's use calculus to denote this.

$$P(q)=\frac{dq}{1}=dq$$

Thus,

$$P(E)={\int }_0^1(\min (2q,1)-q)dq$$

Step 6: Solving the Integral

To solve this integral, we need to split it into two parts:

1. From 0 to 1/2, where $\min (2q,1)=2q$
2. From 1/2 to 1, where $\min (2q,1)=1$

Now,

$P(E)={\int }_0^{1/2}(2q-q)dq+{\int }_{1/2}^1(1-q)dq$

$={\int }_0^{1/2}qdq+{\int }_{1/2}^1(1-q)dq$

$=[\frac{q^2}{2}{]}_0^{1/2}+[q-\frac{q^2}{2}{]}_{1/2}^1$

$=(\frac{1}{8}-0)+((1-\frac{1}{2})-(\frac{1}{2}-\frac{1}{8}))$

$=\frac{1}{8}+\frac{1}{2}-\frac{3}{8}=\frac{1}{4}$

Therefore, the probability that the ratio $p/q$ lies between 1 and 2 is $\frac{1}{4}$ or $0.25$ or $25\%$.

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PS - A similar integration is involved in the problem Random-Angle-II. Check it out for a more detailed explaination incase of any confusions!.