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.

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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?

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  • $\begingroup$ I knew a slightly different result but this seems to be the one I need! Thanks! $\endgroup$ – PhDing Feb 7 '19 at 4:23
  • $\begingroup$ If this hint worked for you, please accept the answer. Thank you! $\endgroup$ – StatsStudent Feb 7 '19 at 15:10
  • $\begingroup$ Sure, I'll try later! $\endgroup$ – PhDing Feb 7 '19 at 15:59
  • $\begingroup$ 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. $\endgroup$ – PhDing Feb 7 '19 at 20:23
  • $\begingroup$ Think about redefining a new variable $Z_i$. Does that help? $\endgroup$ – StatsStudent Feb 7 '19 at 22:20

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