# A segment is removed from a semi circle what is the pdf of the remaining area?

Question: Given the illustration below, the cdf of Z is $$Pr(Z\leq z) = \frac{\text{exp}(-\lambda A_z)-\text{exp}(-N/2)}{1-\text{exp}(-N/2)}$$

How could one Prove that

$$f_z(Z) = \frac{2\lambda \sqrt{R^2-z^2}}{1-\text{exp}(-N/2)} \text{exp}[-\lambda A_z]$$
for $0<z\leq R$

where $A_z$ = $R^2 [\cos^{-1} (z/R)- (z/R)\sqrt{1-(z/R)^2}]$

Challenge: The first problem I have is understanding how $A_z$ was derived, then, how the numerator of the first equation of the pdf was gotten. The closest clue I have to $A_z$ is this question and this one, but here, the height of the segment here is unknown.

In case it might interest someone, $N = \lambda \pi R^2$ and $\lambda$ represents the density of a two dimensional Poisson Point Process

I understand that you have already recognised that this problem can be solved by using the identity relating the PDF of a distribution to the CDF: $f_Z(z) = F_Z'(z)$, which for your particular CDF is equivalent to:

$$f_Z(z) = \frac{-\lambda A'(z) \exp(-\lambda A(z))}{1- \exp(-N/2)}.$$

So it remains to calculate $A(z)$, and then use this to derive $A'(z)$.

Calculating A(z)

In the following I refer to my diagram (unfortunately not as well produced as your graphic, but you should get the point).

Let $\theta_z$ denote the angle made between $A \hat O B$, using the standard trigonometric identify for the cosine we have that

$$\theta_z = \cos^{-1}(z/R).$$

Therefore the area of the circular wedge with angle $2\theta_z$ (i.e. the wedge bounded by the lines $OB$ and $OC$) is

\begin{aligned}W_z &= \pi R^2 \times \frac{2\theta_z}{2 \pi} \\ & = R^2 \theta_z \\ & = R^2 \cos^{-1}(z/R). \end{aligned}

We can now obtain $A(z)$ by subtracting from $W_z$ the area, $T_z$, of the triangle made out by the lines $OBCO$, which is equal to the area of the rectangle with side lengths equal to those of the line segments $OA$ and $AB$.

Since $A$ has coordinates $(z,0)$ the line segment $OA$ has length $z$; we can calculate the coordinates of $B$ by relying on the formula $x^2 + y^2 = R^2$, which gives coordinates $(z, \sqrt{R^2 - z^2})$, and so the length of $AB$ is $\sqrt{R^2 - z^2}$. Combining these we get

$$T_z = z \sqrt{R^2 - z^2}$$,

and further more

\begin{aligned} A(z) &= W_z - T_z \\ & = R^2 \cos^{-1}(z/R) - z \sqrt{R^2 - z^2} \\ & = R^2 \left( \cos^{-1}(z/R) - (z/R) \sqrt{1 - (z/R)^2} \right), \end{aligned} which is as you gave in the question statement.

Calculating $A'(z)$

This now follows from appealing to standard identities for derivatives. If we denote

$$g(x) = \cos^{-1}(x) - x \sqrt{1 - x^2},$$

then we have $A(z) = R^2 g(z/R)$, and so using the chain rule:

\begin{aligned} A'(z) & = R^2 \frac{d}{dz} g(z/R) \\ & = R^2 g'(z/R) \frac{d}{dz} \left(\frac{z}{R}\right) \\ & = R g'(z/R). \end{aligned} So concentrating on $g(x)$ we differentiate this using the identity for the derivative of $\cos^{-1}$, as well as the product rule to handle the second term to obtain:

$$g'(x) = - \frac{1}{\sqrt{1-x^2}} - \frac{1-2x^2}{\sqrt{1-x^2}},$$ where the first term above is contributed from $\cos^{-1}$, and the second is from the product term. Rearranging this we have \begin{aligned} g'(x) &= -2\frac{1 - x^2}{\sqrt{1-x^2}} \\ & = -2 \sqrt{1 - x^2}. \end{aligned} And so we have

\begin{aligned} A'(z) &= Rg'(z/R) \\ & = -2R \sqrt{1 - (z/R)^2} \\ & = -2 \sqrt{R^2 - z^2}. \end{aligned}

Substituting this into the original formula provided for $f_Z(z)$ gives the desired answer.

• Thanks for this brilliant answer. Please could you add some detail on how $A'(z) = R g'(z/R).$ was gotten using the chain rule. I didn't get that part well even with my basic knowledge of the chain rule. – Abdulhameed Jul 25 '18 at 0:09
• I have added a few lines breaking down the chain rule calculation. – owen88 Jul 25 '18 at 14:50
• Those lines are well appreciated and understood. Thank you very much – Abdulhameed Jul 26 '18 at 3:49