Probability Transformation Explanation

The problem describes a random variable X that is obtained by transforming a Uniform(0, 1) random variable U. We need to determine which statements about X are true.

Uniform Variable U

Let U be a random variable following a Uniform distribution on the interval (0, 1). The probability density function (PDF) of U is given by:

\( f_U(u) = \begin{cases} 1 & \text{for } 0 < u < 1 \\ 0 & \text{otherwise} \end{cases} \)

Transformation to Y

The first part of the transformation is \( Y = \pi\left(U - \frac{1}{2}\right) \). Let's find the range of Y:

• Since \( 0 < U < 1 \), we have \( -\frac{1}{2} < U - \frac{1}{2} < 1 - \frac{1}{2} \).
• This gives \( -\frac{1}{2} < U - \frac{1}{2} < \frac{1}{2} \).
• Multiplying by \(\pi\), we get \( -\frac{\pi}{2} < \pi\left(U - \frac{1}{2}\right) < \frac{\pi}{2} \).
• So, the range of Y is \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \).

To find the PDF of Y, we use the transformation method. The relationship is \( y = \pi\left(u - \frac{1}{2}\right) \). The inverse relationship is \( u = \frac{y}{\pi} + \frac{1}{2} \). The derivative of the inverse transformation is \( \left|\frac{du}{dy}\right| = \left|\frac{d}{dy}\left(\frac{y}{\pi} + \frac{1}{2}\right)\right| = \frac{1}{\pi} \). The PDF of Y is:

\( f_Y(y) = f_U(u) \left|\frac{du}{dy}\right| = 1 \cdot \frac{1}{\pi} = \frac{1}{\pi} \quad \text{for } -\frac{\pi}{2} < y < \frac{\pi}{2} \)

This shows that Y follows a Uniform distribution on the interval \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \).

Transformation to X (Cauchy Distribution)

The second part of the transformation is \( X = \tan(Y) \). We need to find the PDF of X. The range of X is obtained by applying the tangent function to the range of Y:

• As \( y \) goes from \( -\frac{\pi}{2} \) to \( \frac{\pi}{2} \), \( x = \tan(y) \) goes from \( -\infty \) to \( \infty \).
• So, the range of X is \( (-\infty, \infty) \).

To find the PDF of X, we use the transformation method. The relationship is \( x = \tan(y) \). The inverse relationship is \( y = \arctan(x) \). The derivative of the inverse transformation is \( \left|\frac{dy}{dx}\right| = \left|\frac{d}{dx}(\arctan(x))\right| = \frac{1}{1+x^2} \). The PDF of X is:

\( f_X(x) = f_Y(y) \left|\frac{dy}{dx}\right| = \frac{1}{\pi} \cdot \frac{1}{1+x^2} \quad \text{for } -\infty < x < \infty \)

This is the PDF of the standard Cauchy distribution, also known as Cauchy(0, 1).

Analyzing Statements about X

Expectation of X⁴

Option 1 states \( E(X^4) = 3 \). The expected value \( E(g(X)) \) is calculated as \( \int_{-\infty}^{\infty} g(x) f_X(x) dx \). For \( g(x) = x^4 \), we need to evaluate \( E(X^4) = \int_{-\infty}^{\infty} x^4 \frac{1}{\pi(1+x^2)} dx \). For this integral to converge, the integrand must decrease fast enough as \(|x| \to \infty\). The integrand behaves like \( \frac{x^4}{x^2} = x^2 \) for large \(|x|\). The integral \( \int_{-\infty}^{\infty} x^2 dx \) diverges. Therefore, the integral \( \int_{-\infty}^{\infty} x^4 \frac{1}{\pi(1+x^2)} dx \) also diverges. \( E(X^4) \) does not exist. Thus, Option 1 is false.

Probability P(X ∈ {1, 2, 5})

Option 2 states \( P(X \in \{1,2,5\})=\frac{1}{2} \). For any continuous random variable X with PDF \( f_X(x) \), the probability of X taking any specific value 'c' is \( P(X=c) = \int_c^c f_X(x) dx = 0 \). The probability of X belonging to a finite set of values \( \{c_1, c_2, ..., c_k\} \) is the sum of probabilities for each value: \( P(X \in \{c_1, ..., c_k\}) = \sum_{i=1}^k P(X=c_i) = \sum_{i=1}^k 0 = 0 \). Therefore, \( P(X \in \{1, 2, 5\}) = 0 \). Thus, Option 2 is false.

Expectation of e^X

Option 3 states \( E(e^X) \) does not exist. We need to evaluate \( E(e^X) = \int_{-\infty}^{\infty} e^x f_X(x) dx = \int_{-\infty}^{\infty} e^x \frac{1}{\pi(1+x^2)} dx \). For this expectation to exist, the integral must converge absolutely. Consider the integral for \( x > 0 \), \( \int_{0}^{\infty} \frac{e^x}{\pi(1+x^2)} dx \). For large \(x\), the term \(e^x\) grows much faster than any polynomial \(1+x^2\). The integral of \(e^x/(1+x^2)\) from 0 to infinity diverges because the integrand does not approach zero fast enough as \(x \to \infty\). Specifically, for \(x \ge 1\), \(1+x^2 \le 2x^2\), so \( \frac{e^x}{1+x^2} \ge \frac{e^x}{2x^2} \). The integral \( \int_1^\infty \frac{e^x}{x^2} dx \) diverges. Since the integral \( \int_{-\infty}^{\infty} e^x \frac{1}{\pi(1+x^2)} dx \) does not converge, \( E(e^X) \) does not exist. Thus, Option 3 is true.

Probability P(X ≤ 0)

Option 4 states \( P(X \leq 0)=\frac{1}{2} \). This is the cumulative distribution function (CDF) of X evaluated at 0. We calculate it using the PDF:

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

This integral is \( \frac{1}{\pi} [\arctan(x)]_{-\infty}^{0} = \frac{1}{\pi} (\arctan(0) - \lim_{z \to -\infty} \arctan(z)) = \frac{1}{\pi} (0 - (-\frac{\pi}{2})) = \frac{1}{\pi} \cdot \frac{\pi}{2} = \frac{1}{2} \). Alternatively, the PDF \( f_X(x) = \frac{1}{\pi(1+x^2)} \) is symmetric about 0, i.e., \( f_X(-x) = f_X(x) \). For a symmetric distribution about 0, the median is 0, and \( P(X \leq 0) = P(X \geq 0) = 0.5 \). Thus, Option 4 is true.

Conclusion

Based on our analysis, statements in Option 3 and Option 4 are true.