Question

The random variable \(x\) has the probability distribution $$ f(x)=e^{-x} \quad(0

Short answer

a) \(\langle x \rangle = 1\), b) \(\langle x_1 + x_2 \rangle = 2\), \(\langle x_1 x_2 \rangle = 1\), c) \( a \sim Gamma(2, 2)\).

Step-by-step solution

- Find the expected value \(\langle x \rangle\)

The expected value of a continuous random variable is calculated as \(\langle x \rangle = \int_0^\infty x f(x) dx\). Given \(f(x) = e^{-x}\) for \(0 < x < \infty\), we substitute and integrate: \[ \langle x \rangle = \int_0^\infty x e^{-x} dx. \]\ This is a standard integral with a result \(\langle x \rangle = 1\).

- Find the expected value \(\langle x_1 + x_2 \rangle\)

For independent variables, the expected value of the sum is the sum of their expected values. So, \[ \langle x_1 + x_2 \rangle = \langle x_1 \rangle + \langle x_2 \rangle = 1 + 1 = 2. \]

- Find the expected value \(\langle x_1 x_2 \rangle\)

For independent variables, the expected value of the product is the product of their expected values. Hence, \[ \langle x_1 x_2 \rangle = \langle x_1 \rangle \cdot \langle x_2 \rangle = 1 \cdot 1 = 1. \]

- Find the probability distribution \(P(a)\) of \(a = \frac{1}{2}(x_{1} + x_{2})\)

Let \(a = \frac{1}{2}(x_{1} + x_{2})\). The moment generating function for \(x_1 + x_2\) is \(M_{x_1 + x_2}(t) = \left(\frac{1}{1-t}\right)^2\) since \(x_i \sim Exp(1)\) and hence \(x_1 + x_2 \sim Gamma(2, 1)\). Now, the moment-generating function for \(a\) is \[ M_a(t) = M_{\frac{1}{2}(x_1 + x_2)}(t) = M_{x_1 + x_2}\left(\frac{t}{2}\right) = \left(\frac{1}{1-t/2}\right)^2. \] Hence the probability distribution \(P(a) \sim Gamma(2, 2)\).

Key concepts

expected value

In probability and statistics, the expected value is a fundamental concept. It's akin to a weighted average of all possible values of a random variable. For a continuous random variable with probability density function (PDF) denoted by \(f(x)\), the expected value (or mean) is given by the integral: \[ \langle x \rangle = \int_0^\infty x \cdot f(x)~dx. \] This formula captures the average outcome if the random process described by the PDF is repeated many times. In our exercise, given \(f(x) = e^{-x}\) for \(0 < x < \infty\), the expected value was calculated as follows: \[ \langle x \rangle = \int_0^\infty x e^{-x}~dx = 1, \] which is a common result for an exponential distribution with a rate parameter of 1.

independent random variables

Two random variables are said to be independent if the occurrence of one does not affect the probability of the occurrence of the other. In mathematical terms, if \(X\) and \(Y\) are independent, then their joint probability distribution is the product of their individual distributions: \[ P(X \le x, Y \le y) = P(X \le x) \cdot P(Y \le y). \] When dealing with expected values of independent random variables, two important properties follow:

• Sum of Expectations: The expected value of the sum of two independent random variables is the sum of their individual expected values, \(\rangle x_1 + x_2 \rangle = \rangle x_1 \rangle + \rangle x_2 \rangle\).
• Product of Expectations: The expected value of the product of two independent random variables is the product of their individual expected values, \(\rangle x_1 \cdot x_2 \rangle = \rangle x_1 \rangle \cdot \rangle x_2 \rangle\).

In the exercise, these properties helped determine:
\( \rangle x_1 + x_2 \rangle = 2 \) and \( \rangle x_1 \cdot x_2 \cdot\rangle = 1. \)

moment-generating function

The moment-generating function (MGF) is a powerful tool in the study of probability distributions. It is defined for a random variable \(X\) as: \ M_X (t) = E(e^{tX}). \ The MGF provides a way to derive moments (i.e., the expected values of powers) of the distribution. If you find the \(n\)-th derivative of \(M_X(t)\) with respect to \(t\) at \(t = 0\), you obtain the \(n\)-th moment, \[ E(X^n) = M_X^{(n)}(0). \] For sums of independent random variables, the MGF of the sum is the product of their individual MGFs. This property was used in the exercise to determine the distribution of \(\frac{1}{2}(x_1 + x_2)\). With \(x_i \sim Exp(1)\), thus \(x_1 + x_2 \sim Gamma(2,1). \) Consequently, the MGF became:
\[ M_{\frac{1}{2}(x_1 + x_2)}(t) = M_{x_1 + x_2}\bigg(\frac{t}{2}\bigg) = \bigg(\frac{1}{1-t/2}\bigg)^2. \]

Gamma distribution

The Gamma distribution is a two-parameter family of continuous probability distributions. The parameters are typically denoted as \(k\) (shape) and \(\theta\) (scale). Its PDF is given by: \[ f(x; k, \theta) = \frac{x^{k-1} e^{-x/\theta}}{\theta^k \Gamma(k)}, \] for \(x > 0\). The Gamma distribution generalizes the exponential distribution. When the shape parameter \(k=1\), it simplifies to the exponential distribution with mean \(\theta\). Common properties include:

• Mean: \(E(X) = k \theta\).
• Variance: \(Var(X) = k \theta^2\).

In the exercise, the sum of two exponential random variables, \(x_1 + x_2 \), followed a \(Gamma(2,1)\) distribution, leading to \(a = \frac{1}{2}(x_1 + x_2) \sim Gamma(2,2)\). This showed the flexibility and comprehensive nature of the gamma distribution in representing different types of data.