Chi-square Random Variable

Definition We say a continuous r.v $\mathbb{X}$ is a Chi-square if it is $gamma( \frac{1}{2}, \frac{k}{2} )$ for some $k > 0$, denoted $\mathbb{X} \sim \chi^2(k)$.

So by the property of $gamma$ distribution, we know:

Property If $\mathbb{X} \sim \chi^2(k)$, $\mu = k, \sigma^2 = \frac{k}{2}$

Theorem If $\mathbb{Z} \sim normal(0, 1)$, then $\mathbb{Z}^2 \sim \chi^2(1)$

Proof $\mathbb{P}(\mathbb{Z}^2 \leq x) = \mathbb{P}(-\sqrt{x} \leq \mathbb{Z} \leq \sqrt{x}) = 2\int_0^{\sqrt{x}} \frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}$ $\text{ so p.d.f } f \text{ of } \mathbb{Z}^2 = \frac{\mathrm{d}}{\mathrm{d}x}2\int_0^{\sqrt{x}} \frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}} = \frac{1}{\sqrt{\pi}}\frac{1}{2}(\frac{1}{2} x)^{-\frac{1}{2}}e^{-\frac{x}{2}} \sim gamma(\frac{1}{2}, \frac{1}{2}) = \chi^2(1)$

Theorem If $\mathbb{Z}_1, ..., \mathbb{Z}_n \sim \text{ i.i.d } normal(0, 1)$, then $\sum_{i=1}^n \mathbb{Z}_i \sim \chi^2(n)$

Proof $M_{Z_i}(t) = (\frac{1/2}{1/2 - t})^{\frac{1}{2}} = (1-2t)^{-\frac{1}{2}}$, so $M_{\sum_{i=0}^n Z_i}(t) = M_{Z_1}(t)^n = (1-2t)^{-\frac{n}{2}} \sim gamma(\frac{1}{2}, \frac{n}{2}) = \chi^2(n)$

lemma If $\mathbb{X} \sim \chi^2(n)$ and $\mathbb{Y} \sim \chi^2(m)$ are independent, then $\mathbb{X}+\mathbb{Y} \sim \chi^2(n+m)$

Example $Z_1 \sim normal(\delta, 1), and Z_2, ..., Z_p \sim normal(0, 1)$ are $p$ independent random variable, then $W = \sum_{i=1}^p Z_i^2 \sim \chi^2_p(\delta^2)$

Solution $V = (Z_1 - \delta)^2 + \sum_{i=2}^p Z_i^2 \sim \chi^2(p)$

$\mathbb{E}[V] = \mathbb{E}[ (Z_1 - \delta)^2 + \sum_{i=2}^p Z_i^2 ] = \mathbb{E}[\sum_{i=1}^p Z_i^2] - 2\delta\mathbb{E}[Z_1] + \delta^2$ $= \mathbb{E}[W] - \delta^2 = p \Rightarrow \mathbb{E}[W] = p + \delta^2$ Now variance $var(V) = var( (Z_1-\delta)^2 ) + 2cov( (Z_1-\delta)^2, \sum_{i=2}^p Z_i^2 ) + var( \sum_{i=2}^p Z_i^2 )$ $= var(Z_1^2 -2\delta Z_1 + \delta^2) + var( \sum_{i=2}^p Z_i^2 )$ $= var(Z_1^2)+2cov(Z_1^2, -2\delta Z_1)+var(-2\delta Z_1) + var( \sum_{i=2}^p Z_i^2 )$ $= var(\sum_{i=1}^p Z_i^2) -4\delta cov(Z_1^2, Z_1) + 4\delta^2 var(Z_1)$ $= var(W) -4\delta(\mathbb{E}[Z_1^3]-\mathbb{E}[Z_1^2]\mathbb{E}[Z_1])+4\delta^2$ Now $\mathbb{E}[Z_1^3]$ can be obtained by m.g.f. Recall that $M_{Z_1}(t) = e^{\delta t + \frac{1}{2}t^2}$ $M_{Z_1}'(t) = (\delta+t)e^{\delta t + \frac{1}{2}t^2}$ $M_{Z_1}''(t) = (\delta+t)^2e^{\delta t + \frac{1}{2}t^2}+e^{\delta t + \frac{1}{2}t^2}$ $M_{Z_1}'''(t) = (\delta+t)^3e^{\delta t + \frac{1}{2}t^2}+3(\delta+t)e^{\delta t + \frac{1}{2}t^2}$ So $M_{Z_1}'''(0) = \delta^3+3\delta$, and $M_{Z_1}''(0) = \delta^2 + 1$. Therefore $var(W) = 2p+4\delta^2$