SOLUTION: The random variables (X,Y) have joint pdf f(x,y) = 1/x for the region 0 < x < 1, 0<y<x.The pdf is zero outside this region. iii. Find the covariance of X, Y i.e. cov(X,Y) iv.

Question 1164373: The random variables (X,Y) have joint pdf f(x,y) = 1/x for the region 0 < x < 1, 0 iii. Find the covariance of X, Y i.e. cov(X,Y)
iv. Find P(X^2+Y^2 ≤ 1| X=x)
v. Find P(X^2+Y^2 ≤ 1)
Answer by CPhill(2264) (Show Source): You can put this solution on YOUR website!
To find the covariance and the probabilities for the joint density function $f(x,y) = \frac{1}{x}$ on the region $0 < x < 1, 0 < y < x$, we proceed with the following evaluations.
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### **iii. Find the Covariance of $X$ and $Y$**
The covariance is defined as:
$$\text{Cov}(X, Y) = E[XY] - E[X]E[Y]$$
**1. Calculate $E[X]$:**
We first find the marginal density $f_X(x)$:
$$f_X(x) = \int_0^x \frac{1}{x} \, dy = \left[ \frac{y}{x} \right]_0^x = 1, \quad 0 < x < 1$$
Now, $E[X] = \int_0^1 x \cdot f_X(x) \, dx = \int_0^1 x \cdot 1 \, dx = \left[ \frac{x^2}{2} \right]_0^1 = \frac{1}{2}$
**2. Calculate $E[Y]$:**
$$E[Y] = \int_0^1 \int_0^x y \cdot \frac{1}{x} \, dy \, dx = \int_0^1 \left[ \frac{y^2}{2x} \right]_0^x \, dx = \int_0^1 \frac{x^2}{2x} \, dx = \int_0^1 \frac{x}{2} \, dx = \left[ \frac{x^2}{4} \right]_0^1 = \frac{1}{4}$$
**3. Calculate $E[XY]$:**
$$E[XY] = \int_0^1 \int_0^x xy \cdot \frac{1}{x} \, dy \, dx = \int_0^1 \int_0^x y \, dy \, dx = \int_0^1 \left[ \frac{y^2}{2} \right]_0^x \, dx = \int_0^1 \frac{x^2}{2} \, dx = \left[ \frac{x^3}{6} \right]_0^1 = \frac{1}{6}$$
**4. Compute Covariance:**
$$\text{Cov}(X, Y) = \frac{1}{6} - \left( \frac{1}{2} \cdot \frac{1}{4} \right) = \frac{1}{6} - \frac{1}{8} = \frac{4 - 3}{24} = \frac{1}{24}$$
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### **iv. Find $P(X^2 + Y^2 \le 1 \mid X = x)$**
Given $X = x$, we look at the conditional density $f(y|x) = \frac{f(x,y)}{f_X(x)} = \frac{1/x}{1} = \frac{1}{x}$ for $0 < y < x$.
The condition $x^2 + y^2 \le 1$ becomes $y^2 \le 1 - x^2$, or **$y \le \sqrt{1 - x^2}$**.
* If $x \le \frac{1}{\sqrt{2}}$, then $x \le \sqrt{1 - x^2}$. The entire range of $y$ (from $0$ to $x$) satisfies the condition. Thus, **$P = 1$**.
* If $x > \frac{1}{\sqrt{2}}$, the condition $y \le \sqrt{1 - x^2}$ is more restrictive than $y < x$.
$$P = \int_0^{\sqrt{1 - x^2}} \frac{1}{x} \, dy = \frac{\sqrt{1 - x^2}}{x}$$
**Result:**
$$P(X^2 + Y^2 \le 1 \mid X = x) =
\begin{cases}
1 & 0 < x \le \frac{1}{\sqrt{2}} \\
\frac{\sqrt{1 - x^2}}{x} & \frac{1}{\sqrt{2}} < x < 1
\end{cases}$$
---
### **v. Find $P(X^2 + Y^2 \le 1)$**
This is the integral of the conditional probability over the marginal density of $X$:
$$P = \int_0^{1/\sqrt{2}} (1)(1) \, dx + \int_{1/\sqrt{2}}^1 \frac{\sqrt{1 - x^2}}{x} (1) \, dx$$
**Calculating the integrals:**
1. **Term 1:** $\int_0^{1/\sqrt{2}} 1 \, dx = \frac{1}{\sqrt{2}}$
2. **Term 2:** Using the substitution $x = \sin\theta$:
$$\int \frac{\cos\theta}{\sin\theta} \cos\theta \, d\theta = \ln| \csc\theta - \cot\theta | + \cos\theta$$
Evaluating from $x = 1/\sqrt{2}$ ($\theta = \pi/4$) to $x = 1$ ($\theta = \pi/2$):
$$[0] - [\ln(\sqrt{2}-1) + 1/\sqrt{2}] = -\ln(\sqrt{2}-1) - \frac{1}{\sqrt{2}}$$
**Total Probability:**
$$P = \frac{1}{\sqrt{2}} - \ln(\sqrt{2}-1) - \frac{1}{\sqrt{2}} = -\ln(\sqrt{2}-1) = \ln\left(\frac{1}{\sqrt{2}-1}\right) = \ln(1+\sqrt{2})$$
$$P \approx 0.8814$$
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