Question

If \(X_{i}, i=1, \ldots, n\) are independent normal random variables, with \(X_{i}\) having mean \(\mu_{i}\) and variance 1, then the random variable \(\sum_{i=1}^{n} X_{i}^{2}\) is said to be a noncentral chi-squared random variable. (a) if \(X\) is a normal random variable having mean \(\mu\) and variance 1 show, for \(|t|<1 / 2\), that the moment generating function of \(X^{2}\) is $$ (1-2 t)^{-1 / 2} e^{\frac{t \mu^{2}}{1-2 t}} $$ (b) Derive the moment generating function of the noncentral chi-squared random variable \(\sum_{i=1}^{n} X_{i}^{2}\), and show that its distribution depends on the sequence of means \(\mu_{1}, \ldots, \mu_{n}\) only through the sum of their squares. As a result, we say that \(\sum_{i=1}^{n} X_{i}^{2}\) is a noncentral chi-squared random variable with parameters \(n\) and \(\theta=\sum_{i=1}^{n} \mu_{i}^{2}\) (c) If all \(\mu_{i}=0\), then \(\sum_{i=1}^{n} X_{i}^{2}\) is called a chi- squared random variable with \(n\) degrees of freedom. Determine, by differentiating its moment generating function, its expected value and variance. (d) Let \(K\) be a Poisson random variable with mean \(\theta / 2\), and suppose that conditional on \(K=k\), the random variable \(W\) has a chi-squared distribution with \(n+2 k\) degrees of freedom. Show, by computing its moment generating function, that \(W\) is a noncentral chi-squared random variable with parameters \(n\) and \(\theta\). (e) Find the expected value and variance of a noncentral chi-squared random variable with parameters \(n\) and \(\theta\).

Short answer

In summary, for a noncentral chi-squared random variable with parameters \(n\) and \(\theta\), the moment generating function is \((1-2t)^{-n/2} e^{\frac{t \theta}{1-2t}}\). Its expected value is \(n+\theta\), and its variance is \(2n+\theta\). Meanwhile, for a chi-squared random variable with \(n\) degrees of freedom, its expected value is \(n\) and its variance is \(2n\).

Step-by-step solution

(a) Deriving the moment generating function of X^2

First we need to find the moment generating function of the random variable \(X\). Recall that the moment generating function of a random variable \(Y\) is defined as: \[ M_Y(t) = E\left[e^{tY}\right] \] For the given normal random variable \(X\) with mean \(\mu\) and variance 1, we can write: \[ M_X(t) = E\left[e^{tX}\right] = \int_{-\infty}^{\infty} e^{tx} \frac{1}{\sqrt{2 \pi}} e^{-\frac{1}{2} (x - \mu)^2}\,dx \] Now we need to find the moment generating function of \(X^2\), which we can write as: \[ M_{X^2}(t) = E\left[e^{tX^2}\right] = \int_{-\infty}^{\infty} e^{tx^2} \frac{1}{\sqrt{2 \pi}} e^{-\frac{1}{2} (x - \mu)^2}\,dx \] Let's use the substitution method: \[ u = x - \mu \] \[ du = dx \] We can rewrite the integral as: \[ M_{X^2}(t) = \int_{-\infty}^{\infty} e^{t(u + \mu)^2} \frac{1}{\sqrt{2 \pi}} e^{-\frac{1}{2} u^2}\,du \] By expanding the exponent term inside the integral and simplifying, we get: \[ M_{X^2}(t) = e^{\frac{t \mu^2}{1 - 2t}} \int_{-\infty}^{\infty} e^{-\frac{1}{2} (1 - 2t)u^2} \frac{1}{\sqrt{2 \pi}}\,du \] With the given condition \( |t| < \frac{1}{2} \), we can further simplify the integral part to: \[ (1-2t)^{-1/2} \] Hence the moment generating function of \(X^2\) is: \[ M_{X^2}(t) = (1-2t)^{-1/2} e^{\frac{t \mu^{2}}{1-2t}} \]

(b) Deriving the moment generating function of the noncentral chi-squared random variable

The given random variable is: \[ Y = \sum_{i=1}^{n} X_i^2 \] where each \(X_i\) is a normal random variable with mean \(\mu_i\) and variance 1. To find the moment generating function of \(Y\), let's first recall the product rule of moment generating functions: If \(Y = Z_1 + Z_2\) then \(M_Y(t) = M_{Z_1}(t) M_{Z_2}(t)\). By applying the product rule to our given random variable \(Y\), we get: \[ M_Y(t) = \prod_{i=1}^{n} M_{X_i^2}(t) = \prod_{i=1}^{n} (1-2t)^{-1/2} e^{\frac{t \mu_i^{2}}{1-2t}} \] And by simplifying this expression, we get: \[ M_Y(t) = (1-2t)^{-n/2} e^{\frac{t \sum_{i=1}^{n}\mu_i^{2}}{1-2t}} \] This shows that the distribution depends on the means \(\mu_1, \ldots, \mu_n\) only through the sum of their squares \(\theta = \sum_{i=1}^{n} \mu_i^2\) and the parameter \(n\).

(c) Finding the expected value and variance of a chi-squared random variable with n degrees of freedom

A chi-squared random variable with \(n\) degrees of freedom has all \(\mu_i = 0\), so the moment generating function of a chi-squared random variable is: \[ M_Y(t) = (1-2t)^{-n/2} \] To find the expected value (mean) and variance, we need to find the first and second moments of the distribution, which is calculated as: \( E[Y] = \frac{d}{dt} M_Y(t) \Big|_{t = 0} \) \( E[Y^2] = \frac{d^2}{dt^2} M_Y(t) \Big|_{t = 0} \) So let's find the first and second derivatives of the moment generating function \(M_Y(t)\): \( \frac{d}{dt} M_Y(t) = -\frac{n}{2} (1-2t)^{-(n+2)/2} (-2) \) \( E[Y] = \frac{d}{dt} M_Y(t) \Big|_{t = 0} = -\frac{n}{2} (1)^{-(n+2)/2}(-2) = n \) Now to calculate the second moment: \( \frac{d^2}{dt^2} M_Y(t) = (-\frac{n}{2})(-\frac{n + 2}{2}) (1-2t)^{-(n+4)/2} (4) \) \( E[Y^2] = \frac{d^2}{dt^2} M_Y(t) \Big|_{t = 0} = (-\frac{n}{2})(-\frac{n + 2}{2}) (1)^{-(n+4)/2}(4) = n(n + 2) \) Now, we can calculate the variance as: \( Var(Y) = E[Y^2] - E[Y]^2 = n(n + 2) - n^2 = 2n \) So the expected value (mean) of a chi-squared random variable with \(n\) degrees of freedom is \(n\) and the variance is \(2n\).

(d) Proving W is a noncentral chi-squared random variable with parameters n and θ

We are given that \(W\) has a conditional chi-squared distribution with \(n+2k\) degrees of freedom, given \(K=k\), and \(K\) is a Poisson random variable with mean \(\frac{\theta}{2}\). We need to find the moment generating function of \(W\). Recall the law of total expectation: \( M_W(t) = E[ E[ e^{tW} | K ]] \) The conditional moment generating function of \(W\) given \(K=k\) can be found as follows: \( E[e^{tW} | K=k] = (1 - 2t)^{-(n + 2k)/2} \) Now, we can find the moment generating function of \(W\): \[ M_W(t) = E[ (1 - 2t)^{-(n + 2K)/2} ] = \sum_{k=0}^{\infty} (1-2t)^{-(n+2k)/2} P(K=k) \] Given that \(K\) follows a Poisson distribution with mean \(\frac{\theta}{2}\), we have: \[ P(K=k) = e^{-\theta/2} \frac{(\theta/2)^k}{k!} \] Plug this into the moment generating function for \(W\): \[ M_W(t) = e^{-\theta/2} \sum_{k=0}^{\infty} \frac{(\theta/2)^k}{k!} (1-2t)^{-(n+2k)/2} \] This is the moment generating function for a noncentral chi-squared random variable with parameters \(n\) and \(\theta\), thus proving \(W\) is a noncentral chi-squared random variable with parameters \(n\) and \(\theta\).

(e) Finding the expected value and variance of a noncentral chi-squared random variable with parameters n and θ

From part (b), we derived the moment generating function of a noncentral chi-squared random variable as: \[ M_Y(t) = (1-2t)^{-n/2} e^{\frac{t \theta}{1-2t}} \] Now let's find the mean and variance by finding the first and second moments: \( E[Y] = \frac{d}{dt} M_Y(t) \Big|_{t = 0} \) \( E[Y^2] = \frac{d^2}{dt^2} M_Y(t) \Big|_{t = 0} \) First, let's find the first derivative of \(M_Y(t)\): \( \frac{d}{dt} M_Y(t) = -\frac{n}{2}(1-2t)^{-(n+2)/2} e^{\frac{t\theta}{1-2t}} + (1-2t)^{-n/2} e^{\frac{t\theta}{1-2t}}\frac{\theta(1-2t)+2t\theta}{(1-2t)^2} \) Now, evaluating the first moment at \(t = 0\): \( E[Y] = -\frac{n}{2}(1)^{-(n+2)/2}(1) + (1)^{-n/2}(1)\frac{\theta}{1} = n + \theta \) Next, let's find the second derivative of \(M_Y(t)\): \( \frac{d^2}{dt^2} M_Y(t) = f_1(t) + f_2(t) \) where \( f_1(t) = (-\frac{n}{2})(-\frac{n+2}{2})(1-2t)^{-(n+4)/2} e^{\frac{t\theta}{1-2t}} \) and \( f_2(t) = (1-2t)^{-n/2} e^{\frac{t\theta}{1-2t}}\frac{(n-2)(1-2t)^3(\theta + 2t\theta)- 2(1-2t)^2(1-2t+\theta)+2t(1-2t)^2}{(1-2t)^6} \) Now evaluating the second moment at \(t = 0\): \( E[Y^2] = f_1(0) + f_2(0) = n(n + 2) + \theta(n + 2\theta) \) Finally, let's find the variance: \( Var(Y) = E[Y^2] - E[Y]^2 = n(n + 2) + \theta(n + 2\theta) - (n + \theta)^2 = 2n + \theta \) So, the expected value of a noncentral chi-squared random variable with parameters \(n\) and \(\theta\) is \(n + \theta\), and its variance is \(2n + \theta\).