I'm following the proof in Cramer's book Mathematical Methods of Statistics, $\S 29.4$. There it is assumed that we have two independent samples $x_1,\ldots, x_{n_1}$ and $y_1,\ldots,y_{n_2}$ drawn from the same normal population $N(\mu,\sigma^2)$. The author claims that we may assume without loss of generality that $\mu = 0$, but I don't see why. I understand that we may shift the random variables by $\mu$ to centralize them, but this is not usually done when calculating this statistic, and I'm trying to figure out why the statistic follows a t-distribution in the general case. I have tried to imitate the proof without the $\mu = 0$ assumption, and the main lines are as follows:

Consider the quadratic form $$Q = n_1 s_1^2 + n_2 s_2^2 = \sum_1^{n_1} x_i^2 + \sum_1^{n_2} y_i^2 - n_1 \overline x^2 - n_2 \overline y^2$$(using uncorrected variances for simplicity), then a transformation partially defined by $z_1 = \sqrt n_1 \overline x$ and $z_2 = \sqrt n_2 \overline y$ and extended to a full orthogonal transformation $(x_1,\ldots,x_{n_1},y_1,\ldots,y_{n_2}) \mapsto (z_1,\ldots,z_{n_1+n_2})$ maps the quadratic form to $$Q = \sum_3^{n_1+n_2} z_i^2,$$ which shows that its rank (the number of degrees of freedom) is $\nu = n_1+n_2-2$. Then define the statistic via the usual formula and do some algebra, $$t = \frac{\overline x - \overline y}{\sqrt{\frac{n_1 s_1^2 + n_2 s_2^2}\nu} \sqrt{\frac1{n_1} + \frac1{n_2}}}= \frac{\sqrt\frac{n_2}{n_1+n_2}z_1 - \sqrt\frac{n_1}{n_1+n_2}z_2}{\sqrt{\frac 1\nu \sum_3^{n_1+n_2} z_i^2}}.$$

One can check that the numerator is a centered distribution, $\sqrt\frac{n_2}{n_1+n_2}z_1 - \sqrt\frac{n_1}{n_1+n_2}z_2 \sim N(0,\sigma^2)$, however one would also hope for the $z_i \sim N(0,\sigma^2)$ in the denominator in order to conclude that this follows a t-distribution with $\nu$ degrees of freedom. Arguing via covariance matrices I see why the orthogonal transformation preserves the variance, but I initially failed to see why the expected value of the $z_i$ $(i \geq 3)$ has to be zero. I finally managed to prove it by noting the following observation:

An orthogonal matrix $C$ of size $n_1+n_2$ whose first two rows are $\frac1{\sqrt n_1}(1,1,\ldots,1,0,0,\ldots0,0)$ ($n_1$ nonzero entries) and $\frac1{\sqrt n_2}(0,0,\ldots,0,1,1,\ldots,1)$ ($n_2$ nonzero entries) has the property that the coefficients in each other row sum to zero. Proof: this equivalently says that $C$ should map the vector $(1,1,\ldots,1)$ to $(\sqrt n_1, \sqrt n_2,0,0,\ldots,0)$. But this is true since $C$ must preserve the norm, as $\|(1,1,\ldots,1)\| = n_1+n_2 = \|(\sqrt n_1, \sqrt n_2, ?,?,\ldots,?)\|$ forces all other entries to vanish.

In our situation, this forces $E(z_i) = \sum_j C_{ij}E(x_j) = \mu \sum_j C_{ij} = 0$ whenever $i \geq 3$ no matter what $\mu$ is. But this seems a bit convoluted.

In summary, my question is:
Is there a more straightforward approach to understanding why the statistic defined above follows a t-distribution even in the case $\mu \neq 0$, or why the author claims that $\mu = 0$ may be assumed without loss of generality? It seems quite nontrivial to me.


  • $\begingroup$ Under the null, when you take the difference of means in the numerator the common $\mu$ cancels out from the distribution of the statistic; $\mu$ has no effect on the statistic at all. $\endgroup$
    – Glen_b
    May 30, 2022 at 1:38
  • $\begingroup$ @Glen_b I agree for the numerator, but don't we also have to ensure that the normal distributions in the denominator also have zero expected value? By definition (still following the book), a t-distribution with $\nu$ d.o.f. is given by a ratio $\frac{\xi_0}{\sqrt{\frac1{\nu} \sum_1^{\nu} \xi_i^2}}$ with all the $\xi_i$ independent and normal $(0,\sigma)$. If only the numerator has zero expected value but the variables I'm squaring on the bottom don't, then I don't think I get a t-distribution, do I? $\endgroup$ May 30, 2022 at 13:41
  • $\begingroup$ Can you explain in more detail why you think $\mu$ comes into the denominator at all? $\endgroup$
    – Glen_b
    Jun 1, 2022 at 0:55
  • $\begingroup$ @Glen_b Because the denominator consists of variables $z_i$ defined nonconstructively (although we can use e.g. Gram–Schmidt) by transforming the variables $x_i$ and $y_i$, and the latter don't satisfy $\mu = 0$. My whole question basically boils down to "why can we assume that $\mu$ doesn't come into the denominator?". I think I managed to prove it via the orthogonal matrix argument, but I can't help but feel like I'm perhaps missing something obvious. It seems like for you it's obvious that $\mu$ doesn't come into the denominator; if you could explain why that would answer my question! $\endgroup$ Jun 2, 2022 at 2:05
  • $\begingroup$ I guess I really should say "non-canonically" instead of "nonconstructively"; my points is that the $z_i$ are obtained through any orthogonal transformation whose first two rows are fixed. (And I believe I proved that this implies that all the other rows sum to zero, hence that the $\mu = 0$.) $\endgroup$ Jun 2, 2022 at 2:13

1 Answer 1


The simple answer is that the statistic does not depend at all on $\mu$, and this is much easier to see from the original, non-transformed formula:

$$t = \frac{\overline x - \overline y}{\sqrt{\frac{n_1 s_1^2 + n_2 s_2^2}\nu} \sqrt{\frac1{n_1} + \frac1{n_2}}}.$$

Indeed under the transformations $x_i \mapsto x_i - \mu$ and $y_i \mapsto y_i - \mu$, we have $\overline x \mapsto \overline x - \mu$ and $\overline y \mapsto \overline y - \mu$, and also $s_1^2 = \frac1n \sum (x_i - \overline x)^2 \mapsto s_1^2$ and similarly $s_2^2 \mapsto s_2^2$, so that the variable $t$ is invariant under horizontal shifts of the parent distribution. This is why we can assume $\mu = 0$ without loss of generality.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.