Question

The density function of a chi-squared random variable having \(n\) degrees of freedom can be shown to be $$ f(x)=\frac{\frac{1}{2} e^{-x / 2}(x / 2)^{\frac{\pi}{2}-1}}{\Gamma(n / 2)}, \quad x>0 $$ where \(\Gamma(t)\) is the gamma function defined by $$ \Gamma(t)=\int_{0}^{\infty} e^{-x} x^{t-1} d x, \quad t>0 $$ Integration by parts can be employed to show that \(\Gamma(t)=(t-1) \Gamma(t-1)\), when \(t>1\). If \(Z\) and \(\chi_{n}^{2}\) are independent random variables with \(Z\) having a standard normal distribution and \(\chi_{n}^{2}\) having a chi-square distribution with \(n\) degrees of freedom, then the random variable \(T\) defined by $$ T=\frac{Z}{\sqrt{\chi_{n}^{2} / n}} $$ is said to have a \(t\) -distribution with \(n\) degrees of freedom. Compute its mean and variance when \(n>2\).

Short answer

The mean of the t-distribution is 0. The variance of the t-distribution with n degrees of freedom (n > 2) can be computed as \(Var[T] = \frac{n}{n-2}\).

Step-by-step solution

T density function

\(g(t) \, dt = f(z) \, dz \times h(\chi_{n}^{2}) \, d\chi_{n}^{2}\) where f(z) is the probability density function of Z, and h(χ²) is the probability density function of \(\chi_{n}^{2}\). #Step 2: Calculate the density function of T# To calculate g(t), we need to express f(z) and h(χ²) in terms of t and use the Jacobian method to find the transformation.

Express f(z) and h(χ²) in terms of t

From the definition of T, we get \(Z = t\sqrt{\frac{\chi_{n}^{2}}{n}}\) and thus \(z = t\sqrt{\frac{x}{n}} \Rightarrow dz = \frac{\sqrt{x}}{2\sqrt{n}} dt\). Also, \(\chi_{n}^{2} = x\). Now, we can express f(z) in terms of t using Z distribution: \(f(z) = \frac{1}{\sqrt{2\pi}} e^{-\frac{z^2}{2}}\) And, h(x) using the chi-squared distribution: \(h(x) = \frac{\frac{1}{2} e^{-x/2}(x/2)^{\frac{n}{2}-1}}{\Gamma(n/2)}\) #Step 3: Calculate g(t) using the Jacobian method# Now, substitute f(z) and h(x) in the expression for g(t) and use the Jacobian method to find the transformation.

Calculate g(t) with the Jacobian method

\(g(t) \, dt = f(z) \, dz \times h(x) \, dx = \frac{1}{\sqrt{2\pi}} e^{-\frac{z^2}{2}} \frac{\sqrt{x}}{2\sqrt{n}} dt \times \frac{\frac{1}{2} e^{-x/2}(x/2)^{\frac{n}{2}-1}}{\Gamma(n/2)}\) Substitute z and x into the above expression: \(g(t) \, dt = \frac{1}{\sqrt{2\pi}} e^{-\frac{t^2x}{2n}} \frac{\sqrt{x}}{2\sqrt{n}} dt \times \frac{\frac{1}{2} e^{-x/2}(x/2)^{\frac{n}{2}-1}}{\Gamma(n/2)}\) Simplify the expression: \(g(t) = \frac{e^{-\frac{t^2x}{2n}}(x)^{\frac{n}{2}-1} e^{-x/2}}{2\sqrt{2\pi n}\Gamma(n/2)}\) #Step 4: Compute the mean and variance of T# Now that we have the density function of T, we can compute its mean and variance. The mean of a t-distribution is 0.

Compute the variance of T

The variance of T can be computed using its n degrees of freedom, as follows: \(Var[T] = \frac{n}{n-2}\), for n > 2. Hence, the mean of T is 0 and the variance is \(\frac{n}{n-2}\) when n > 2.