Question

Let \(X_{1}\) and \(X_{2}\) be independent random variables with uniform distribution over the interval \(\left[\theta-\frac{1}{2}, \theta+\frac{1}{2}\right] .\) Show that \(X_{1}-X_{2}\) has a distribution independent of \(\theta\) and find its density function.

Short answer

The random variable \(Y = X_1 - X_2\) has a probability density function given by: \(f_Y(y) = \begin{cases} 1 & \text{for } y \in [-1, 1] \\ 0 & \text{otherwise} \end{cases}\). This distribution is independent of the parameter \(\theta\) and is a uniform distribution on the interval \([-1, 1]\).

Step-by-step solution

Find the Probability Density Functions of \(X_1\) and \(X_2\)

The random variables \(X_1\) and \(X_2\) both have a uniform distribution over the interval \(\left[\theta-\frac{1}{2}, \theta+\frac{1}{2}\right]\). The probability density function (pdf) of a continuous uniform distribution over an interval \([a, b]\) is given by: \(f(x) = \frac{1}{b-a}\) for \(x \in [a, b]\) and \(f(x) = 0\) otherwise. Therefore, the pdfs of \(X_1\) and \(X_2\) are the same and can be written as: $f_{X_1}(x_1) = f_{X_2}(x_2) = \begin{cases} 1 & \text{for } x_1, x_2 \in \left[\theta-\frac{1}{2}, \theta+\frac{1}{2}\right] \\ 0 & \text{otherwise} \end{cases}$

Compute the Probability Density Function of \(Y\) using the Convolution Theorem

To find the probability density function of \(Y = X_1 - X_2\), we use the convolution theorem: \(f_{Y}(y) = \int_{-\infty}^{\infty} f_{X_1}(x_1) f_{X_2}(x_1 - y) dx_1\) However, the functions \(f_{X_1}(x_1)\) and \(f_{X_2}(x_1-y)\) are non-zero only within the interval \(\left[\theta-\frac{1}{2}, \theta+\frac{1}{2}\right]\). Hence we need to consider the convolution integral only within these bounds: \(f_{Y}(y) = \int_{\theta-\frac{1}{2}}^{\theta+\frac{1}{2}} f_{X_1}(x_1) f_{X_2}(x_1 - y) dx_1\) Since both \(f_{X_1}(x_1)\) and \(f_{X_2}(x_1-y)\) are equal to 1 within the given intervals, the convolution integral simplifies to: \(f_Y(y) = \int_{\theta-\frac{1}{2}}^{\theta+\frac{1}{2}} dx_1\)

Evaluate the Convolution Integral and Verify Independence of \(\theta\)

The convolution integral is just the integral of 1 over \(x_1\) in the interval \(\left[\theta-\frac{1}{2}, \theta+\frac{1}{2}\right]\): \(f_Y(y) = \left(x_1\right) \Big|_{\theta-\frac{1}{2}}^{\theta+\frac{1}{2}} = (\theta+\frac{1}{2}) - (\theta-\frac{1}{2}) = 1\) As we can see, the result \(f_Y(y) = 1\) does not depend on the parameter \(\theta\). Hence, the distribution of \(Y = X_1 - X_2\) is independent of \(\theta\).

Find the Range of \(Y\)

To find the range of \(Y\), we need to consider the minimum and maximum values that \(Y = X_1 - X_2\) can take. From the range of \(X_1\) and \(X_2\), we have: Minimum of \(Y\): \(X_1 - X_2 = (\theta - \frac{1}{2}) - (\theta + \frac{1}{2}) = -1\) Maximum of \(Y\): \(X_1 - X_2 = (\theta + \frac{1}{2}) - (\theta - \frac{1}{2}) = 1\) Hence, the range of \(Y\) is \([-1, 1]\). Finally, we find the density function of \(Y\) as follows: $f_Y(y) = \begin{cases} 1 & \text{for } y \in [-1, 1] \\ 0 & \text{otherwise} \end{cases}$ This is our final result, showing that the random variable \(Y = X_1 - X_2\) has a uniform distribution over [-1, 1], independent of the parameter \(\theta\).