The Noncentral T Distribution Under a Shifted Mean

Why a Shifted Mean Changes the t Law

Studentization does not remove a nonzero difference between the population mean and a reference value. It scales that difference by the standard error, while retaining random variation in both the sample mean and the estimated standard deviation. The resulting statistic therefore has a noncentral t distribution, rather than the central t distribution obtained when the reference value equals the true mean.

The key parameters are the degrees of freedom, determined by the sample variance, and the noncentrality parameter, which measures the mean difference in standard-error units using the population standard deviation. The representation as a shifted normal divided by an independent chi-square scale makes the distribution precise and provides a direct route to its properties.

Definition From a Normal Sample

Let \(X_1,\ldots,X_n\) be independent and identically distributed as \(N(\mu,\sigma^2)\), where \(n\geq 2\) and \(\sigma>0\). Write \(\bar X=n^{-1}\sum_{i=1}^n X_i\) and \(S^2=(n-1)^{-1}\sum_{i=1}^n(X_i-\bar X)^2\). For a fixed reference value \(\mu_0\), consider \(T=(\bar X-\mu_0)/(S/\sqrt n)\).

Definition: A random variable \(T\) has a noncentral t distribution with \(\nu\) degrees of freedom and noncentrality parameter \(\delta\), written \(T\sim t_\nu(\delta)\), if \(T=(Z+\delta)/\sqrt{U/\nu}\), where \(Z\sim N(0,1)\), \(U\sim\chi^2_\nu\), and \(Z\) and \(U\) are independent. Here \(\nu>0\) and \(\delta\in\mathbb R\). In the normal-sample setting, \(\nu=n-1\) and \(\delta=\sqrt n(\mu-\mu_0)/\sigma\).

The representation follows from the normal sampling results for the sample mean and sample variance. Specifically, \(Z=\sqrt n(\bar X-\mu)/\sigma\) is standard normal, \(U=(n-1)S^2/\sigma^2\) has a chi-square distribution with \(n-1\) degrees of freedom, and these two random variables are independent. Splitting the numerator into its true-mean fluctuation and its fixed offset gives

T =\frac{\sqrt n(\bar X-\mu_0)}{S} =\frac{\sqrt n(\bar X-\mu)/\sigma+\sqrt n(\mu-\mu_0)/\sigma} {\sqrt{(n-1)S^2/\sigma^2\,/\,(n-1)}} =\frac{Z+\delta}{\sqrt{U/\nu}}, \qquad \nu=n-1,\quad \delta=\frac{\sqrt n(\mu-\mu_0)}{\sigma}.

Thus the denominator has the same distribution whether or not the mean equals \(\mu_0\), but the numerator has a normal shift \(\delta\). If \(\mu=\mu_0\), then \(\delta=0\) and this is the central t distribution. A positive or negative \(\delta\) records the direction as well as the standardized size of the difference.

Worked Example: Finding the Noncentrality Parameter

Suppose \(n=16\), \(\sigma=2\), the true mean is \(\mu=10.5\), and the statistic is formed relative to \(\mu_0=10\). Determine its exact sampling distribution.

The degrees of freedom are \(\nu=n-1=15\). The noncentrality parameter is

\delta=\frac{\sqrt{16}(10.5-10)}{2} =\frac{4(0.5)}{2}=1.

The normal-sample representation therefore gives \(T=(Z+1)/\sqrt{U/15}\), with \(Z\sim N(0,1)\), \(U\sim\chi^2_{15}\), and \(Z\) independent of \(U\). Hence \(T\sim t_{15}(1)\). The parameter is positive because the true mean exceeds the reference value.

Distribution Function and Density

The ratio representation gives an exact cumulative distribution function by conditioning on \(U\). The chi-square denominator is positive with probability one, so for each \(u>0\), the event \(T\leq t\), conditional on \(U=u\), is equivalent to \(Z\leq t\sqrt{u/\nu}-\delta\). Let \(\Phi\) denote the standard normal cumulative distribution function.

Theorem: If \(T\sim t_\nu(\delta)\), then for every real \(t\),

F_{\nu,\delta}(t) =P(T\leq t) =E\!\left[\Phi\!\left(t\sqrt{U/\nu}-\delta\right)\right], \qquad U\sim\chi^2_\nu.

If \(\nu>0\), its density is

f_{\nu,\delta}(t) =\int_0^\infty \sqrt{\frac{u}{\nu}}\, \phi\!\left(t\sqrt{\frac{u}{\nu}}-\delta\right) \frac{u^{\nu/2-1}e^{-u/2}}{2^{\nu/2}\Gamma(\nu/2)} \,du, \qquad -\infty<t<\infty,

where \(\phi\) is the standard normal density and \(\Gamma\) is the gamma function.

For the cumulative distribution function, condition on \(U\) and use the standard normal law of \(Z\):

P(T\leq t\mid U) =P\!\left(Z+\delta\leq t\sqrt{U/\nu}\mid U\right) =\Phi\!\left(t\sqrt{U/\nu}-\delta\right).

Taking expectations over \(U\) proves the stated formula. For the density, condition on \(U=u\). The transformation from \(Z\) to \(T\) is \(t=(z+\delta)/\sqrt{u/\nu}\), or \(z=t\sqrt{u/\nu}-\delta\), with derivative \(dz/dt=\sqrt{u/\nu}\). Therefore the conditional density of \(T\) is \(\sqrt{u/\nu}\,\phi(t\sqrt{u/\nu}-\delta)\). Multiplying by the chi-square density of \(U\) and integrating over \(u>0\) gives the displayed density. This mixture argument also ensures it integrates to one: each conditional density integrates to one, and the chi-square density integrates to one. \(\square\)

The density is generally expressed as an integral, rather than as a single elementary formula. The integral is already useful: it shows how the random denominator mixes normal densities with different scales. Setting \(\delta=0\) recovers the central t density.

Worked Example: An Exact Probability About the Sign

Suppose \(n=25\) and the population mean exceeds the reference value by \(0.4\) when \(\sigma=2\). Find the probability that the studentized statistic is positive.

The noncentrality parameter is

\delta=\frac{\sqrt{25}(0.4)}{2} =\frac{5(0.4)}{2}=1.

Because \(\sqrt{U/\nu}>0\) with probability one, \(T>0\) if and only if \(Z+\delta>0\). Consequently,

P(T>0)=P(Z>-1)=\Phi(1)\approx 0.8413.

The degrees of freedom are \(\nu=24\), but they do not enter this particular sign probability: a positive denominator cannot change the sign of the numerator. The corresponding probability of a negative statistic is \(P(T<0)=\Phi(-1)\approx 0.1587\).

Mean and Variance

The noncentrality parameter shifts the numerator but does not eliminate the variability introduced by the random scale. This can be quantified by deriving the moments. The chi-square density gives, for any \(r>-\nu/2\),

E[U^r] =\frac{1}{2^{\nu/2}\Gamma(\nu/2)} \int_0^\infty u^{r+\nu/2-1}e^{-u/2}\,du =2^r\frac{\Gamma(\nu/2+r)}{\Gamma(\nu/2)}.

The last equality follows by substituting \(v=u/2\) into the integral defining \(\Gamma(\nu/2+r)\). In particular, \(E[U^{-1/2}]\) is finite when \(\nu>1\), and \(E[U^{-1}]\) is finite when \(\nu>2\).

Theorem: If \(T\sim t_\nu(\delta)\), then for \(\nu>1\),

E[T]=\delta\sqrt{\frac{\nu}{2}}\, \frac{\Gamma((\nu-1)/2)}{\Gamma(\nu/2)}.

For \(\nu>2\),

E[T^2]=(1+\delta^2)\frac{\nu}{\nu-2}, \qquad \operatorname{Var}(T) =(1+\delta^2)\frac{\nu}{\nu-2} -\delta^2\frac{\nu}{2} \left(\frac{\Gamma((\nu-1)/2)}{\Gamma(\nu/2)}\right)^2.

To prove the mean formula, use the independent representation \(T=(Z+\delta)\sqrt{\nu/U}\). When \(\nu>1\), \(E[U^{-1/2}]\) is finite, so independence and \(E[Z+\delta]=\delta\) yield

E[T] =\delta\sqrt{\nu}\,E[U^{-1/2}] =\delta\sqrt{\frac{\nu}{2}}\, \frac{\Gamma((\nu-1)/2)}{\Gamma(\nu/2)}.

For the second moment, independence, \(E[(Z+\delta)^2]=1+\delta^2\), and \(E[U^{-1}]=1/(\nu-2)\) give

E[T^2] =\nu E[(Z+\delta)^2]E[U^{-1}] =(1+\delta^2)\frac{\nu}{\nu-2}.

Subtracting the square of the mean gives the variance formula. The conditions matter: the mean is not finite for \(\nu\leq1\), and the second moment, hence the variance, is not finite for \(\nu\leq2\). \(\square\)

Worked Example: Moments With Three Degrees of Freedom

Let \(n=4\), \(\sigma=4\), and \(\mu-\mu_0=1\). Then \(\nu=3\) and \(\delta=\sqrt{4}(1)/4=1/2\). Since \(\nu>2\), both the mean and variance exist.

Using \(\Gamma(1)=1\) and \(\Gamma(3/2)=\sqrt{\pi}/2\), the mean is

E[T] =\frac12\sqrt{\frac32}\, \frac{\Gamma(1)}{\Gamma(3/2)} =\frac12\sqrt{\frac32}\frac{2}{\sqrt{\pi}} =\sqrt{\frac{3}{2\pi}} \approx 0.691.

The second moment and variance are

E[T^2]=(1+1/4)\frac{3}{3-2}=\frac{15}{4}=3.75, \qquad \operatorname{Var}(T)=\frac{15}{4}-\frac{3}{2\pi} \approx 3.273.

The mean is not simply \(\delta=0.5\): the random denominator changes the mean as well as the spread. In this case its variability is substantial, as the second moment confirms.

Interpretation and a Common Mistake

The noncentrality parameter \(\delta=\sqrt n(\mu-\mu_0)/\sigma\) describes the mean difference relative to the population standard error \(\sigma/\sqrt n\). For a fixed standardized difference, increasing the sample size increases \(|\delta|\), because the mean is more precisely estimated. This is why noncentral t distributions arise in exact power calculations: under an alternative mean, the test statistic follows a noncentral rather than central t law.

The sign also has a simple distributional interpretation. Since the denominator is positive, \(P(T>0)=P(Z>-\delta)=\Phi(\delta)\). Further, replacing \(Z\) by \(-Z\) shows that \(-T\) under parameter \(\delta\) has the same distribution as \(T\) under parameter \(-\delta\). For \(\delta\ne0\), the distribution is generally not symmetric about zero.

Mistake: treating the result as a location shift of a central t variable. One might write \(T= T_0+\delta\), where \(T_0=Z/\sqrt{U/\nu}\). But the actual representation is \(T=T_0+\delta/\sqrt{U/\nu}\). The shift is divided by the same random scale, so it is not a fixed additive constant.
Correction: retain the full ratio \((Z+\delta)/\sqrt{U/\nu}\). Both the normal numerator and the chi-square denominator matter to the law.
Mistake: using the central t law whenever the data are normal. Normal sampling supplies the independent normal and chi-square components, but the central t distribution requires the reference value to equal the true mean. A shifted mean gives \(\delta\ne0\).

Key takeaway: For a normal sample standardized against a fixed reference mean, the exact statistic has distribution \(t_{n-1}(\delta)\), where \(\delta=\sqrt n(\mu-\mu_0)/\sigma\). The shift changes the numerator while the independent chi-square scale remains random.

Check Your Understanding

Use the ratio representation and the stated moment conditions to answer these questions.

For a sample of size \(n=36\), what are the degrees of freedom of the noncentral t statistic?
If \(\mu=\mu_0\), what value does the noncentrality parameter take, and what distribution results?
For \(T\sim t_\nu(\delta)\), explain why \(P(T>0)=\Phi(\delta)\), regardless of \(\nu\).
For which values of \(\nu\) does the mean exist? For which values does the variance exist?
Why is \(T\) not in general equal in distribution to a central t random variable plus the constant \(\delta\)?