# How to find the CDF and PDF [closed]

How to find the Cumulative Distribution Function (CDF) and Probability Distribution Function (PDF) for uniform variable inside circle given by $R^2 = (X-c_1)^2+(Y-c_2)^2$, where ($c_1, c_2$) is the center of circle.

I think that the PDF for problem is given by

$$f(x,y) = \begin{cases} \frac{1}{\pi R^2}&, \quad (x-c_1)^2 + (y-c_2)^2 \leq R^2\\ 0&, \quad otherwise \end{cases}$$

• What is the variable? – GoodDeeds Sep 18 '16 at 5:32
• Sorry, the variable is $T = \sqrt{X^2+Y^2}$ – Wagner Jorge Sep 18 '16 at 5:33
• What is CDF, PDF? – user261263 Sep 18 '16 at 5:36
• I edited the question. – Wagner Jorge Sep 18 '16 at 5:39
• @WagnerJorge Put that in the opening post. $\uparrow$ – Graham Kemp Sep 18 '16 at 5:57

HINT $$\begin{split} F_T(t) &= \mathbb{P}[T < t] \\ &= \mathbb{P}\left[X^2 + Y^2 < t^2\right] \\ &= \frac{1}{(b-a)^2} \int_a^b \int_a^b \mathbb{I}_{x^2 + y^2 < t^2}dxdy \end{split}$$ and you can transform the integral over the indicator by integrating over the correct region...
• Applying polar coordinates the integral above is given by $F_T(t) = \frac{1}{(b-a)^2}\int_{0}^{t}\int_{0}^{2\pi}rd\theta dr$, then the cumulative distribution function is $F_T(t) = \frac{\pi t^2}{(b-a)^2}$. Correct? – Wagner Jorge Sep 18 '16 at 6:22