# Showing a Normal and a Chi square are independent

Student's t distribution is defined as the ratio of a standard normally distributed random variable and the square root of a Chi-square distributed random variable divided by its degrees of freedom, given that they are independent. In formulas one can write $$\frac{Z}{\sqrt \frac{U}{df}}$$, where $$Z$$ is $$N(0,1)$$ and $$U$$ is $$\chi^2_{df}$$.

In showing that this statement is true, I arrived at the point in which I have $$\frac{(n-1)S^2}{\sigma^2}\sim\chi^2_{n-1}$$ and $$\frac{\bar{X}-\mu}{\frac{\sigma}{\sqrt n}}\sim N(0,1)$$. Then, following the definition, we would have that $$\begin{gather} \frac{\frac{\bar{X}-\mu}{\frac{\sigma}{\sqrt n}}}{\sqrt\frac{\frac{(n-1)S^2}{\sigma^2}}{n-1}} \end{gather}$$

is distributed as a $$t_{n-1}$$. But I am stuck at how to prove than this two random variables are independent between them.We covered a result about independence in the case of two Chi-square random variables and I thought of seeing the standard Normal as the square of a Chi-square random variable but I am afraid of it being mathematically sacrilegious.

Do you have any hint?

This is a brute force solution requiring just multivariable calculus.

It suffices to prove that the sample mean $$\bar{X} = \frac{1}{n} \sum_{i=1}^n X_i$$ and the sample variance $$S^2 = \frac{1}{n - 1} \sum_{i=1}^n \left(X_i - \bar{X}\right)^2$$ are independent. Thus, it suffices to prove that the sample mean $$\bar{X}$$ is independent of the vector $$(X_1 - \bar{X}, \ldots, X_n - \bar{X}).$$ Moreover, since \begin{aligned} \sum_{i=1}^n (X_i - \bar{X}) &= \sum_{i=1}^n X_i - \sum_{i=1}^n \bar{X} \\ &= n \bar{X} - n \bar{X} \\ &= 0, \end{aligned} and hence $$X_1 - \bar{X} = -\sum_{i=2}^n (X_2 - \bar{X}),$$ it follows that $$X_1 - \bar{X}$$ can be recovered from just knowing $$(X_2 - \bar{X}, \ldots, X_n - \bar{X})$$.

Thus, it suffices to prove that the sample mean $$\bar{X}$$ is independent from $$(X_2 - \bar{X}, \ldots, X_n - \bar{X}).$$

Now consider the joint density \begin{aligned} f_{(X_1, \ldots, X_n)}(x_1, \ldots, x_n) &= \left(2 \pi \sigma^2\right)^{-n/2} \exp\left(-\sum_{i=1}^n \frac{1}{2}\left(\frac{x_i - \mu}{\sigma}\right)^2\right) \\ &= \left(2 \pi \sigma^2\right)^{-n/2} \exp\left(-\sum_{i=1}^n \frac{1}{2}\left(\frac{x_i - \bar{x}}{\sigma}\right)^2 - \frac{n}{2}\left(\frac{\bar{x} - \mu}{\sigma}\right)^2\right) \\ &= \underbrace{\left(2 \pi \sigma^2\right)^{-n/2}}_{\text{constant}} \underbrace{\exp\left(-\sum_{i=1}^n \frac{1}{2}\left(\frac{x_i - \bar{x}}{\sigma}\right)^2\right)}_{\text{depends only on (x_2-\bar{x},\ldots,x_n-\bar{x})}} \underbrace{\exp\left(-\frac{n}{2}\left(\frac{\bar{x} - \mu}{\sigma}\right)^2\right)}_{\text{depends only on \bar{x}}}. \end{aligned} To get from $$(X_1,\ldots,X_n)$$ to $$(\bar{X}, X_2 - \bar{X}, \ldots, X_n - \bar{X})$$, consider the diffeomorphism $$T : \mathbb{R}^n \to \mathbb{R}^n$$ given by $$T(x_1, \ldots, x_n) = (\bar{x}, x_2 - \bar{x}, \ldots, x_n - \bar{x}).$$ ($$T$$ is a diffeomorphism since it's clearly differentiable and its inverse is given by $$T^{-1}(y_1, \ldots, y_n) = \left(n y_1 - \sum_{i=2}^n y_i, y_2 + y_1, \ldots, y_n + y_1\right),$$ which is also clearly differentiable). Up to transpose, the Jacobian matrix of $$T$$ is $$DT(x_1, \ldots, x_n) = \begin{bmatrix} 1/n & 1/n & 1/n & \cdots & 1/n \\ -1/n & (n - 1) / n & -1/n & \cdots & -1/n \\ -1/n & -1/n & (n - 1) / n & \cdots & -1/n \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ -1/n & -1/n & -1/n & \cdots & (n - 1)/n. \end{bmatrix},$$ which doesn't depend on $$x_1, \ldots, x_n$$. Thus, the determinant of $$DT$$ is some constant $$C$$. Now the joint density of $$(\bar{X}, X_2 - \bar{X}, \ldots, X_n - \bar{X})$$ satisfies $$f_{(\bar{X}, X_2 - \bar{X}, \ldots, X_n - \bar{X})}(y_1, \ldots, y_n) = |C| f_{(X_1, \ldots, X_n)}(T^{-1}(y_1, \ldots, y_n))$$ which factors as a function of $$y_1$$ times a function of $$(y_2, \ldots, y_n)$$ by what was shown above.

Therefore, $$\bar{X}$$ and $$(X_2 - \bar{X}, \ldots, X_n - \bar{X})$$ are independent.

I'll provide a hint to your self-study question: A corollary of a classic statistical theorem states that if $$\mathbf{x} \sim N_p(\boldsymbol{\mu}, \sigma^2\boldsymbol{I})$$,then $$\mathbf{Bx}$$ and $$\mathbf{x^\prime Ax}$$ are independent if and only if $$\mathbf{BA}$$ is equal to the zero matrix. So, perhaps you could write the numerator as $$\mathbf{Bx}$$ and the denominator as $$\mathbf{x^\prime Ax}$$ and work from there?

• I knew a slightly different result but this seems to be the one I need! Thanks! Feb 7, 2019 at 4:23
• If this hint worked for you, please accept the answer. Thank you! Feb 7, 2019 at 15:10
• Sure, I'll try later! Feb 7, 2019 at 15:59
• I am fine with the denominator, which can be rewritten as $\frac{X-\mu e}{\sigma}'\frac{ee'}{n-1}\frac{X-\mu e}{\sigma}$, and has the required form. I am stil stuck with the numerator because it is already a Normal, then the only thing I can think of is multiplying by an identity matrix but this is not working. Feb 7, 2019 at 20:23
• Think about redefining a new variable $Z_i$. Does that help? Feb 7, 2019 at 22:20