Random Angle II

Now we must find $P(\theta >\pi /3)$.

Note that,

$\tan (\theta )=\frac{A}{B}$

where $A$ and $B$ both are IID Uniform(0,1) random variables.

For convenience, let's fix $B$ for now. Then, this problem becomes much simpler. It reduces to:

Given $\tan (\theta )=\frac{A}{B}$, where $A$ is Uniform(0,1) ,$B$ is a given constant,
find $P(\theta >\pi /3)$.

Let's try and solve this sub-problem:

$P(\theta >\pi /3)=P(\tan (\theta )>\sqrt{3})=P(\frac{A}{B}>\sqrt{3})=P(A>B\sqrt{3})$

Now, since $A$ is a Uniform(0,1) random variable, clearly:

$P(A>x)=\max (0,1-x)$

(How? 🤔)

• If $x$ is between 0 and 1 , well and good.
• If $x>1$, since A cannot be greater than 1, $P(A>x)=0$.
• Still confused? Try intuitively finding $P(A>0)$, $P(A>1/2)$, and $P(A>1)$. You will see the essence of the above expression.
• Still confused? Read up on PDF (Probability Density Function) and CDF (Cumulative Density Function)
  of the continuous Uniform distribution .

Therefore, for the smaller sub-problem above:

$P(A>B\sqrt{3})=1-B\sqrt{3}$

for all $1-B\sqrt{3}\ge 0\Rightarrow B\le 1/\sqrt{3}$ .

Essentially, this is equivalent to writing:

$P(\theta >\pi /3\mid B)=1-B\sqrt{3}$

(Why? 🤔)
Because $P(\theta >\pi /3\mid B)$ represents the probability of $\theta$ being greater than $\pi /3$,
given some B / conditioned on some B .

This is exactly what we calculated above! By fixing $B$ , we found $P(\theta >\pi /3)$
in terms of some $B$ , for a given $B$.

But of course, as the question states, $B$ is also a Uniform(0,1) random variable.

We shall now use the fundamental law of total probability here.

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Step 2: Using the Law of Total Probability

Note that we can say:

$P(\theta >\pi /3)=\sum P(\theta >\pi /3\mid B)P(B)$

for all $B$.

Now, we must compute $P(B)$ .

What exactly does this represent? 🤔

It basically asks:

What is the probability of choosing a particular $B$ from $[0,1]$?

Very small, right?

How do we quantify this? Using calculus!

We can say that the probability of choosing a particular $B$ from $[0,1]$ is essentially:

$\frac{dB}{1}=dB$

where $dB$ is a very small term.

Notice that now our summation actually reduces to a simple integration , due to our usage of
"elemental B" or "$dB$" to model $P(B)$.

Ideally, we should vary $B$ from $0$ to $1$, but note that $B$ must be ≤ $\frac{1}{\sqrt{3}}$ ,
otherwise $P(\theta >\pi /3\mid B)=0$ .

Therefore, we rewrite the above expression as a definite integral:

$P(\theta >\pi /3)={\int }_0^{1/\sqrt{3}}(1-B\sqrt{3})dB$

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Step 3: Solving the Integral

Expanding the integral:

${\int }_0^{1/\sqrt{3}}(1-B\sqrt{3})dB$

We solve separately:

$\int 1dB=B$

$\int B\sqrt{3}dB=\frac{\sqrt{3}{B}^2}{2}$

Evaluating from $0$ to $1/\surd 3$:

$P(\theta >\pi /3)={\left[B-\frac{\sqrt{3}{B}^2}{2}\right]}_0^{1/\sqrt{3}}$

Substituting $B=1/\surd 3$:

$P(\theta >\pi /3)=\frac{1}{\sqrt{3}}-\frac{\sqrt{3}(1/3)}{2}$

$=\frac{1}{\sqrt{3}}-\frac{\sqrt{3}}{6}$

$=\frac{2}{2\sqrt{3}}-\frac{1}{2\sqrt{3}}$

$=\frac{1}{2\sqrt{3}}$

Thus,

$\boxed{P(\theta >\pi /3)=\frac{1}{2\sqrt{3}}}$

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Conclusion

Using a combination of probability concepts, the law of total probability, and calculus ,
we found that the probability of $\theta$ exceeding $\frac{\pi }{3}$ is $\frac{1}{2\surd 3}$ .

This problem highlights how fundamental probability techniques can be used to analyze geometric randomness .