Linear combination $C=\sum_{i=1}^{n} a_i \bar{X}_i$ is called a (estimated) contrast if $\sum_{i=1}^{n} a_i=0$. Two contrasts are called orthogonal if $\sum_{i=1}^{n} a_i b_i = 0$; simplest example would be $C_1=\bar{X}_1 - \bar{X}_2$ and $C_2=\bar{X}_1 + \bar{X}_2 - 2\bar{X}_3$. I have read the claim

Two orthogonal contrasts are statistically independent

in Statistical Design and Analysis of Experiments by Robert L. Mason (2003, 2nd edition, Page 248) . I have searched the web for a proof with no success. I have tried to prove $$f\left( \sum_{i=1}^{n} a_i \bar{X}_i = v_2 \mid \sum_{j=1}^{n} b_j \bar{X}_j = v_1\right) = f\left(\sum_{i=1}^{n} a_i \bar{X_i} = v_2\right) \tag 1 $$ or even uncorrelatedness (a weaker property) $$\mathbb{E}\left [\left(\sum_{i=1}^{n} a_i \bar{X}_i\right)\left(\sum_{j=1}^{n} b_j \bar{X}_j\right)\right ]=\mathbb{E}\left[\sum_{i=1}^{n} a_i \bar{X}_i\right]\mathbb{E}\left[\sum_{j=1}^{n} b_j \bar{X}_j\right] \tag 2$$ again with no success. It seems there are more intricacies involved which I am missing.


I have managed to prove the uncorrelatedness $(2)$ using kjetil b halvorsen's advice. For now, let's assume there is only one factor, then $Y_{il} = \mu_{i} + \epsilon_{il}; \epsilon_{il} \overset{\text{iid}}{\sim} \mathcal{N}(0, \sigma^2)$ is the assumed model for i-th factor-level and l-th repeat test. Now let $\bar{X}_i \triangleq \frac{1}{r}\sum_{l=1}^r Y_{il}$, where $r$ is the number of repeat tests in a balanced experiment. Therefore, $$\begin{align*} &\mathbb{E}\left [\left(\sum_{i=1}^{n} a_i \bar{X}_i\right)\left(\sum_{j=1}^{n} b_j \bar{X}_j\right)\right ]=\sum_{i,j=1}^n a_ib_j \mathbb{E}\left [\bar{X}_i\bar{X}_j \right] \\ &\overset{\bar{X}_i \perp \bar{X}_j \text{ for } i \neq j}{=} \sum_{i\neq j} a_ib_j \mu_i \mu_j + \sum_{i=1}^n a_ib_i \mathbb{E}\left [\bar{X}_i^2 \right] \\ \end{align*}$$ By using $$\begin{align*} \mathbb{E}\left [\bar{X}_i^2 \right] &= \mathbb{E}\left [\left(\frac{1}{r}\sum_{l=1}^r Y_{il}\right) \left(\frac{1}{r}\sum_{l'=1}^r Y_{il'}\right)\right] = \frac{1}{r^2}\sum_{l,l'=1}^r\mathbb{E}\left [Y_{il} Y_{il'}\right] \\ &\overset{Y_{il} \perp Y_{il'} \text{ for } l \neq l'}{=} \frac{1}{r^2}\left(\sum_{l\neq l'} \mathbb{E}[Y_{il}]\mathbb{E}[Y_{il'}] + \sum_{l=1}^r \mathbb{E}\left [Y_{il}^2 \right]\right) \\ &= \frac{1}{r^2}\left( \sum_{l\neq l'} \mu_i^2 + \sum_{l=1}^r \mathbb{E}\left [(\mu_i + \epsilon_{il})^2 \right] \right) \\ &= \frac{1}{r^2}\left( r(r-1)\mu_i^2 + \sum_{l=1}^r \left (\mu_i ^ 2 + \sigma ^2 \right) \right) = \mu_i^2 + \frac{1}{r} \sigma^2 \end{align*}$$ We arrive at $$\begin{align*} &\mathbb{E}\left [\left(\sum_{i=1}^{n} a_i \bar{X}_i\right)\left(\sum_{j=1}^{n} b_j \bar{X}_j\right)\right ]= \sum_{i,j=1}^n a_ib_j \mu_i \mu_j + \frac{\sigma^2}{r}\sum_{i=1}^n a_ib_i \\ &\overset{\sum_{i=1}^n a_ib_i=0}{=} \sum_{i,j=1}^n a_ib_j \mathbb{E}[\bar{X}_i] \mathbb{E}[\bar{X}_j] = \mathbb{E}\left[\sum_{i=1}^{n} a_i \bar{X}_i\right]\mathbb{E}\left[\sum_{j=1}^{n} b_j \bar{X}_j\right] \end{align*}$$ Extending this proof to arbitrary $n$ factors and averages is straightforward; it can be messy though. For example, for a 3-factor experiment, any average $\bar{X}_{i'}$ can be expressed as $\bar{X}_{i'} \triangleq \sum_{ijkl} a_{ijk}Y_{ijkl}$, which can be replaced in the proof.

There is only one question left. Is "statistically independent" is used only loosely to mean "uncorrelated"? Or $(1)$ and $(2)$ can be proved to be equivalent in this context?

  • 1
    $\begingroup$ This needs the normal assumption on the errors, and using that, showing correlation is zero is enough. also, you need to define your notation $\bar{X}_i$. Are these averages based on equal number of observations, say? $\endgroup$ Commented Jun 15, 2021 at 15:20

2 Answers 2


The proof of uncorrelatedness can be done in a more straightforward fashion: Assume that $\bar X_i, i=1,\dots,n$ are uncorrelated. Let $$ C_1 = \sum_i a_i \bar X_i, \quad C_2 = \sum_i b_i \bar X_i, \quad $$ We have \begin{align*} \text{cov}(C_1, C_2) &= \text{cov}(\sum_i a_i \bar X_i, \sum_j b_j X_j) \\ &= \sum_{i} \sum_j a_i b_j \text{cov}(\bar X_i, \bar X_j) \\ &= \sum_{i} a_i b_i \text{var}(\bar X_i). \end{align*} The first equation is by the bilinrarity of the covariance operator. The second equality is by the cross-terms vanishing due to uncorrelatedness. Hence, $C_1$ and $C_2$ are uncorrelated if and only if $ \sum_{i} a_i b_i \text{var}(\bar X_i) = 0$.

Regarding the second question: If $\bar X_i, i = 1,\dots,n$ are (jointly) Gaussian, then uncorrelatedness is equivalent to independence. This is a well-known property of multivariate Gaussian distribution. Otherwise, in general, independence does not follow from uncorrelatedness:

For example, assume that $\bar X_1$ and $\bar X_2$ are independent and both uniformly distributed in $[-1,1]$. Consider $C_1 = \bar X_1 + \bar X_2$ and $C_2 = \bar X_1 - \bar X_2$. Then $C_1$ and $C_2$ will be uncorrelated but dependent. To see why they are dependent, you can plot the support of the density of $(C_1,C_2)$ in the 2-D plan which would be a rotated square (a diamond). Alternatively, note that if $C_1 = 2$, then we should have $\bar X_1 = \bar X_2 =1$, hence $C_2 = 0$. That is, the distribution of $C_2$ conditioned on $C_1 = 2$ is a point mass at $0$. On the other hand, argue that the distribution of $C_2$ conditioned on $C_1 = 0$ will be uniform is $[-2,2]$ (the picture helps here). This shows that they are dependent.

  • $\begingroup$ Very nice! Thanks! $\endgroup$
    – Esmailian
    Commented Jun 16, 2021 at 5:46
  • 1
    $\begingroup$ I added some comments about your second question. $\endgroup$
    – passerby51
    Commented Jun 16, 2021 at 13:22

I managed to prove the statement thanks to passerby51 for hinting the jointly Gaussian distribution. The proof goes as follows:

  1. Show averages in each contrast are independent
  2. Show contrasts are uncorrelated
  3. Use Theorem 1: nonsingular transformation of independent random variables $X_1,\dots,X_n$, where $X_i \sim \mathcal{N}(0, \sigma_i^2)$, is jointly Gaussian
  4. Use Theorem 2: jointly Gaussian random variables that are uncorrelated are independent

I am assuming (a) a fixed-effect ANOVA model $Y_{\alpha l} = \mu_{\alpha} + \epsilon_{\alpha l}; \epsilon_{\alpha l} \overset{\text{iid}}{\sim} \mathcal{N}(0, \sigma^2)$ where $\alpha$ denotes factor combination and $l$ denotes repeat test, and (b) experiment is balanced ($r$ repeat tests per combination). Accordingly, an average can be expressed as $\bar{X}_i \triangleq \frac{1}{r|I|}\sum_{\alpha \in I, l}Y_{f(i,\alpha) l}$. For example, in a 3-factor experiment, an average would be $\bar{X}_j \triangleq \frac{1}{acr}\sum_{ikl} Y_{ijkl}$ where $\alpha=ik$, $f(j,\alpha)=ijk$, and $I=\left\{(i, k) \mid i \in [1, a], k \in [1, c] \right\}$.

  1. For any two averages to be independent, I assume they do not share any $\epsilon_{\alpha l}$, i.e. each average $i$ contains an exclusive set of factor-level combinations $f(i, I)$. For example, in a 2-factor balanced experiment, for $\bar{X}_i \triangleq \frac{1}{br}\sum_{jl} Y_{ijl}$ (average response for factor-level $i$), any $\bar{X}_i$ and $\bar{X}_j$ do not share a $\epsilon_{\alpha l}$ with the same index, thus $\bar{X}_i \perp \bar{X}_j$ for $i \neq j$.

  2. As passerby51 neatly showed, two contrasts are uncorrelated iff $\sum_i a_ib_i\text{Var}(\bar{X}_i) = 0$. Now by using representation $\bar{X}_i = \frac{1}{r|I|}\sum_{\alpha \in I, l}Y_{f(i,\alpha) l}$ for averages, we have $$\text{Var}(\bar{X}_i) = \frac{1}{r^2|I|^2}\sum_{\alpha \in I,l}\sigma^2=\frac{\sigma^2}{r|I|} \Rightarrow \sum_i a_ib_i\text{Var}(\bar{X}_i) = \frac{\sigma^2}{r|I|}\sum_i a_ib_i \overset{\sum_{i=1}^n a_ib_i=0}{=} 0$$

  3. We express contrasts $C_1 = \sum_{i=1}^{n} a_i \bar{X}_i$ and $C_2 = \sum_{j=1}^{n} b_j \bar{X}_j$ in a matrix form as follows $$\begin{align*} \begin{bmatrix} C_1\\ C_2\\ \dots \end{bmatrix} &= \begin{bmatrix} a_1& \dots & a_n\\ b_1& \dots & b_n\\ . & \dots & . \end{bmatrix} \left( \begin{bmatrix} \bar{X}_1 - \mathbb{E}[\bar{X}_1] \\ \bar{X}_2 - \mathbb{E}[\bar{X}_2]\\ \dots \end{bmatrix} + \begin{bmatrix} \mathbb{E}[\bar{X}_1] \\ \mathbb{E}[\bar{X}_2] \\ \dots \end{bmatrix} \right) \\ &\Rightarrow \mathbf{C} = A\mathbf{X} + A\mathbf{\mu} = A\mathbf{X} +\mathbf{\mu'} \end{align*}$$ where $\mathbb{E}[\bar{X}_i]=\frac{1}{|I|}\sum_{\alpha \in I}\mu_{f(i,\alpha)}$, and $X_i \triangleq \bar{X}_i - \mathbb{E}[\bar{X}_i] = \frac{1}{r|I|}\sum_{\alpha \in I, l}\epsilon_{f(i,\alpha) l} \overset{\text{iid}}{\sim} \mathcal{N}(0, \sigma^2/r|I|)$. The other rows of matrix $A$ (3rd row and below) are filled with dummy values just to have $n$ linearly independent rows, thus a nonsingular $A$; here orthogonality of contrasts came into play. Now using the fact that jointly Gaussian distribution is closed under marginalization of the rest of (dummy) contrasts, Theorem 1 implies that $C_1$ and $C_2$ are jointly Gaussian.

  4. Using Theorem 2, statistical independence of two orthogonal contrasts follows immediately.

Theorems 1 and 2 can be found in this Lecture by Prof. Robert B. Ash


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