Probability Density Function Analysis

The question provides the probability density function (PDF) for a continuous random variable X:

\(f(x)=\frac{1}{\pi} \frac{1}{1+(x+1)^2}\), for \(-\infty < x < \infty\).

This is the PDF of a Cauchy distribution centered at \(x_0 = -1\) with a scale parameter \(\gamma = 1\).

A new random variable Y is defined based on X:

\(Y=\left\{\begin{array}{cc} \frac{X}{|X|}, & \text { if } X \neq 0 \\ 0, & \text { if } X=0 \end{array}\right.\)

Let's analyze the possible values of Y:

• If \(X > 0\), then \(|X| = X\), so \(Y = \frac{X}{X} = 1\).
• If \(X < 0\), then \(|X| = -X\), so \(Y = \frac{X}{-X} = -1\).
• If \(X = 0\), then \(Y = 0\).

Since X is a continuous random variable, the probability of X being exactly 0 is \(P(X=0) = 0\). Therefore, \(P(Y=0) = 0\).

The random variable Y can only take values -1 or 1 (with non-zero probability).

Calculating Probabilities for Y

The probability that \(Y=1\) is the probability that \(X > 0\). We calculate this by integrating the PDF of X from 0 to \(\infty\):

\(P(Y=1) = P(X > 0) = \int_{0}^{\infty} f(x) dx = \int_{0}^{\infty} \frac{1}{\pi (1+(x+1)^2)} dx\)

Let \(u = x+1\). Then \(du = dx\). When \(x=0\), \(u=1\). When \(x=\infty\), \(u=\infty\).

\(P(X > 0) = \frac{1}{\pi} \int_{1}^{\infty} \frac{1}{1+u^2} du\)

The integral of \(\frac{1}{1+u^2}\) is \(\arctan(u)\).

\(P(X > 0) = \frac{1}{\pi} [\arctan(u)]_{1}^{\infty} = \frac{1}{\pi} (\lim_{u \to \infty} \arctan(u) - \arctan(1))\)

\(P(X > 0) = \frac{1}{\pi} (\frac{\pi}{2} - \frac{\pi}{4}) = \frac{1}{\pi} (\frac{\pi}{4}) = \frac{1}{4}\)

So, \(P(Y > 0) = P(Y=1) = \frac{1}{4}\).

The probability that \(Y=-1\) is the probability that \(X < 0\). We calculate this by integrating the PDF of X from \(-\infty\) to 0:

\(P(Y=-1) = P(X < 0) = \int_{-\infty}^{0} f(x) dx = \int_{-\infty}^{0} \frac{1}{\pi (1+(x+1)^2)} dx\)

Let \(u = x+1\). Then \(du = dx\). When \(x=-\infty\), \(u=-\infty\). When \(x=0\), \(u=1\).

\(P(X < 0) = \frac{1}{\pi} \int_{-\infty}^{1} \frac{1}{1+u^2} du\)

\(P(X < 0) = \frac{1}{\pi} [\arctan(u)]_{-\infty}^{1} = \frac{1}{\pi} (\arctan(1) - \lim_{u \to -\infty} \arctan(u))\)

\(P(X < 0) = \frac{1}{\pi} (\frac{\pi}{4} - (-\frac{\pi}{2})) = \frac{1}{\pi} (\frac{\pi}{4} + \frac{2\pi}{4}) = \frac{1}{\pi} (\frac{3\pi}{4}) = \frac{3}{4}\)

So, \(P(Y < 0) = P(Y=-1) = \frac{3}{4}\).

The probability distribution of Y is:

Evaluating the Statements

Statement 1: E(Y) = 0

The expected value of Y is calculated as:

\(E(Y) = \sum y \cdot P(Y=y)\)

\(E(Y) = (-1) \cdot P(Y=-1) + (0) \cdot P(Y=0) + (1) \cdot P(Y=1)\)

\(E(Y) = (-1) \cdot \frac{3}{4} + (0) \cdot 0 + (1) \cdot \frac{1}{4}\)

\(E(Y) = -\frac{3}{4} + 0 + \frac{1}{4} = -\frac{2}{4} = -\frac{1}{2}\)

Since \(E(Y) = -1/2 \neq 0\), statement 1 is false.

Statement 2: P(Y > 0) < P(Y < 0)

We found \(P(Y > 0) = P(Y=1) = \frac{1}{4}\) and \(P(Y < 0) = P(Y=-1) = \frac{3}{4}\).

The statement is \(\frac{1}{4} < \frac{3}{4}\). This inequality is true.

So, statement 2 is true.

Statement 3: P(Y < -1) < P(Y > 1)

The possible values for Y are -1, 0, and 1.

\(P(Y < -1)\): Y cannot take any value less than -1. Therefore, \(P(Y < -1) = 0\).

\(P(Y > 1)\): Y cannot take any value greater than 1. Therefore, \(P(Y > 1) = 0\).

The statement is \(0 < 0\). This inequality is false, as \(0 = 0\).

So, statement 3 is false.

Statement 4: E(Y²) = 1

Let's consider the random variable \(Y^2\).

• If \(Y = -1\), \(Y^2 = (-1)^2 = 1\).
• If \(Y = 0\), \(Y^2 = 0^2 = 0\).
• If \(Y = 1\), \(Y^2 = 1^2 = 1\).

So, \(Y^2\) can take values 0 or 1.

The probability \(P(Y^2 = 0)\) is \(P(Y = 0)\), which is 0.

The probability \(P(Y^2 = 1)\) is \(P(Y = -1 \text{ or } Y = 1) = P(Y = -1) + P(Y = 1)\).

\(P(Y^2 = 1) = \frac{3}{4} + \frac{1}{4} = 1\).

So, \(Y^2\) is a discrete random variable that takes the value 1 with probability 1.

The expected value of \(Y^2\) is:

\(E(Y^2) = \sum y^2 \cdot P(Y^2=y^2)\)

\(E(Y^2) = (0) \cdot P(Y^2=0) + (1) \cdot P(Y^2=1)\)

\(E(Y^2) = (0) \cdot 0 + (1) \cdot 1 = 0 + 1 = 1\).

Since \(E(Y^2) = 1\), statement 4 is true.

Conclusion

Based on our analysis, statements 2 and 4 are true.

• Statement 1: \(E(Y) = 0\) (False, \(E(Y) = -1/2\))
• Statement 2: \(P(Y > 0) < P(Y < 0)\) (True, \(1/4 < 3/4\))
• Statement 3: \(P(Y < -1) < P(Y > 1)\) (False, \(0 = 0\))
• Statement 4: \(E(Y^2) = 1\) (True)