probabilityprobability distributions
Let $X$ and $Y$ be jointly distributed random
variables having the joint probability
density function
[
f(x,y) = \begin{cases}
\frac{1}{\pi}, &\text{if}\quad x^2 + y^2 \leq 1\\
0, &\text{otherwise}\\
\end{cases}
Then $P(Y > |X|)$ is
I tried writing integration for this but I am getting wrong. I tried writing cdf for this function as well but that did not help. How should I proceed further?
If I were you, I would use transformations.
$$ Y1 = X1 - X2 = u_1(x_1,x_2)$$ $$ Y2 = X1 + X2 = u_2(x_1,x_2)$$ $$ X_1 = \frac{Y_1+Y_2}{2}=w_1(y_1,y_2)$$ $$ X_2 = \frac{Y_1-Y_2}{2}=w_2(y_1,y_2)$$
$$ f(y_1,y_2) = f(w_1(y1, y2),w_2(y1,y2)|J|$$
j is jaconian
I solved same problem with standard normal distributions. I added picture of my process. or You can solve this problem by using distribution method.
good luck :)
By independence, the joint density function is $f_{X,Y}(x,y) = f_X(x) f_Y(y)$. Be sure to incorporate the information about the support of $f_X$ and $f_Y$ (where it is zero and where it is nonzero).
Similarly, the joint distribution function is $P(X\le x, Y \le y) = P(X \le x) P(Y \le y)$ by independence. Again, be careful: $P(X\le x)$ is piecewise linear.
Best Answer
A figure indicating the range of $(X, Y)$ often helps to determine the integral bounds and solve problems like this.
In this problem, the range of $(X, Y)$ is given by $0 \leq x^2 + y^2 \leq 1$, which is the unit circle (showing below).
Since the area of the unit circle is $\pi \times 1^2 = \pi$ and given that $$f(x,y) = \frac{1}{\pi} \quad \text{if } x^2 + y^2 \leq 1,$$ and 0 otherwise, we conclude that $(X, Y)$ is uniformly distributed over the unit circle (think about why this is true).
Next, we want to find $P(Y > \left|X\right|)$. Since $(X, Y)$ is uniformly distributed, the problem reduces to finding the ratio between areas of the region defined by the event $\{Y > \left|X\right|\}$ (i.e. highlighted in light blue) and the unit circle. From the figure we can see that the highlighted region is in fact a quarter of the unit circle (the angle between $y = x$ and $y = -x$ is 90 degrees), we have $$P(Y > \left|X\right|) = \frac{\frac{1}{4}\pi}{\pi} = \frac{1}{4}.$$
Remark: This approach makes use of the property of uniform distribution, which bypasses calculating the integral. If we do want to solve this problem using the integral approach, note that \begin{align*} P(Y > \left|X\right|) &= \iint_{\{y > \left|x\right|\}} f(x,y) dxdy\\ &= \iint_{\{y > \left|x\right|, \ x^2 + y^2 \leq 1\}} \frac{1}{\pi} dxdy\\ &= \frac{1}{\pi} \iint_{\{y > \left|x\right|, \ x^2 + y^2 \leq 1\}} dxdy\\ &= \frac{1}{\pi} \int_{\pi/4}^{3\pi/4} \int_0^1 rdrd\theta \quad (\text{polar coordinate})\\ &= \frac{1}{\pi} \int_{\pi/4}^{3\pi/4} \left[\frac{1}{2}r^2\right]\bigg|_0^1 d\theta\\ &= \frac{1}{\pi} \int_{\pi/4}^{3\pi/4} \frac{1}{2} d\theta\\ &= \frac{1}{2\pi} \left[\frac{3\pi}{4} - \frac{\pi}{4}\right]\\ &= \frac{1}{2\pi} \cdot \frac{\pi}{2}\\ &= \frac{1}{4}. \end{align*} More details on polar coordinate transformation can be found here.