The joint probability density function (PDF) is given by $f(x, y) = \frac{1}{\pi}$ for $0 \leq x^2 + y^2 \leq 1$, which describes a uniform distribution over the unit disk centered at the origin. The total area of the unit disk is $\pi$. We need to evaluate the given statements.

Analyzing Statement 1: Independence

For $X$ and $Y$ to be independent, their joint PDF must equal the product of their marginal PDFs, i.e., $f(x, y) = f_X(x) f_Y(y)$.

• Marginal PDF of X: $f_X(x) = \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \frac{1}{\pi} dy = \frac{1}{\pi} [y]_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} = \frac{2\sqrt{1-x^2}}{\pi}$ for $-1 \leq x \leq 1$.
• Marginal PDF of Y: $f_Y(y) = \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} \frac{1}{\pi} dx = \frac{1}{\pi} [x]_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} = \frac{2\sqrt{1-y^2}}{\pi}$ for $-1 \leq y \leq 1$.
• Product of marginal PDFs: $f_X(x) f_Y(y) = \left(\frac{2\sqrt{1-x^2}}{\pi}\right) \left(\frac{2\sqrt{1-y^2}}{\pi}\right) = \frac{4\sqrt{(1-x^2)(1-y^2)}}{\pi^2}$.

Since $f_X(x) f_Y(y) \neq f(x, y)$, the random variables $X$ and $Y$ are not independent.

Analyzing Statement 2: $P(X > 0)$

The probability $P(X > 0)$ corresponds to the integral of the PDF over the region where $x > 0$ within the unit disk. Since the PDF is uniform ($1/\pi$) over the unit disk (area $\pi$), the probability is calculated as:

$P(X > 0) = \iint_{x^2+y^2 \leq 1, x > 0} f(x, y) dx dy$

Geometrically, the region $x > 0$ within the unit disk is the right half of the disk. This region has half the area of the unit disk.

Probability = (Area of the region) $\times$ (Density)

$P(X > 0) = (\frac{1}{2} \times \text{Area of Unit Disk}) \times \frac{1}{\pi} = (\frac{1}{2} \times \pi) \times \frac{1}{\pi} = \frac{1}{2}$.

Therefore, the statement $P(X > 0) = 1/2$ is correct.

Analyzing Statement 3: $E(Y)$

The expected value of $Y$ is calculated as:

$E(Y) = \iint_{-\infty}^{\infty} y f(x, y) dx dy = \iint_{x^2+y^2 \leq 1} y \left(\frac{1}{\pi}\right) dx dy$

The region of integration, the unit disk, is symmetric with respect to the x-axis (y=0). The integrand $y \left(\frac{1}{\pi}\right)$ is an odd function of $y$. The integral of an odd function over a symmetric interval is zero.

Explicitly, $E(Y) = \int_{-1}^{1} \left( \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \frac{y}{\pi} dy \right) dx$. The inner integral is $\frac{1}{\pi} [\frac{y^2}{2}]_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} = 0$. Thus, $E(Y) = \int_{-1}^{1} 0 dx = 0$.

Therefore, the statement $E(Y) = 0$ is correct.

Analyzing Statement 4: $Cov(X, Y)$

The covariance is defined as $Cov(X, Y) = E(XY) - E(X)E(Y)$.

• We already found $E(Y) = 0$. By symmetry of the unit disk and the PDF with respect to the y-axis, $E(X)$ must also be 0.
• Now, calculate $E(XY)$: $E(XY) = \iint_{-\infty}^{\infty} xy f(x, y) dx dy = \iint_{x^2+y^2 \leq 1} xy \left(\frac{1}{\pi}\right) dx dy$.
• The integrand $xy \left(\frac{1}{\pi}\right)$ is an odd function with respect to $x$ (for fixed $y$) and an odd function with respect to $y$ (for fixed $x$). The region of integration (unit disk) is symmetric about both the x-axis and y-axis.
• Therefore, the integral $E(XY)$ evaluates to 0. Explicitly, $E(XY) = \int_{-1}^{1} \left( \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \frac{xy}{\pi} dy \right) dx$. The inner integral is $\frac{x}{\pi} [\frac{y^2}{2}]_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} = 0$. Thus, $E(XY) = \int_{-1}^{1} 0 dx = 0$.
• Covariance calculation: $Cov(X, Y) = E(XY) - E(X)E(Y) = 0 - (0)(0) = 0$.

Therefore, the statement $Cov(X, Y) = 0$ is correct.

Conclusion

Statements B ($P(X > 0) = 1/2$), C ($E(Y) = 0$), and D ($Cov(X, Y) = 0$) are correct. Statement A ($X$ and $Y$ are independent) is incorrect.