Define $\bar{\bar{Y}}=\sum n_i \bar{Y}_{i.}/\sum n_i$ and $\bar{\theta}=\sum n_i\theta_i / \sum n_i$, where $Y_i \sim N(\theta,\sigma^2)$. How does can I show that $\frac{1}{\sigma^2}\sum^k_{i=1}n_i[(\bar{Y}_{i.}-\bar{\bar{Y}})-(\theta_i-\bar{\theta)}]^2 \sim \chi^2_{k-1}$ under ANOVA assumptions?

my work:

Let $\bar{U}_i=\bar{Y}_{i.}-\theta_i$, for $i=1,...,n$. So, $\bar{U}_i \sim N(0,\frac{\sigma^2}{n_i})$.

Let $\bar{\bar{U}}=\bar{\bar{Y}}-\bar{\theta}$. So, $\bar{\bar{U}} \sim N(0,\frac{\sigma^2}{\sum n_i})$.

The linear combination $\bar{U}_i-\bar{\bar{U}} \sim N(0,\sigma^2(\frac{1}{n_i}+\frac{1}{\sum n_i}))$.

Hence we can rewrite the original expression as $\frac{1}{\sigma^2}\sum^k_{i=1}n_i(\bar{U}_{i}-\bar{\bar{U}})^2$.

I feel like my work is close, but I messed up somewhere, and I need help finding my error. Due to my variance term of the distribution given by $\bar{U}_{i}-\bar{\bar{U}}$, the given expression does not appear to follow a $\chi^2_{k-1}$ distribution. Where did I mess up?

updated work:

I have that $\frac{\sum n_i \bar{U}_i^2}{\sigma^2}=\frac{\sum n_i (\bar{U}_i-\bar{\bar{U}})^2}{\sigma^2}+\frac{\sum n_i \bar{\bar{U}}^2}{\sigma^2}$, where $\frac{\sum n_i \bar{U}_i^2}{\sigma^2} \sim \chi^2_k$ and $\frac{\sum n_i \bar{\bar{U}}^2}{\sigma^2}\sim \chi^2_1$

However, I need to show that the two added terms on the right-hand side of the inequality are independent. How may I go about this?

  • 1
    $\begingroup$ $\bar U_i$ and $\bar{\bar U}$ are not independent. // For simplicity, maybe begin with a balanced design with all $n_i = n.$ Then the sample mean of the $k$ group means $\bar U_i$ is $\bar{\bar U}.$ For a random sample $X_1, \dots, X_n,$ from $\mathsf{Norm}(\mu, \sigma),$ it is known that $\frac{(n-1)S^2}{\sigma^2} \sim \mathsf{Chisq}(\nu = n-1).$ $\endgroup$
    – BruceET
    May 4, 2020 at 22:56
  • $\begingroup$ @BruceET I see. I made a mistake in assuming that $\bar{U}_i$ and $\bar{\bar{U}}$ were independent. Since $\bar{U}_i \sim N(0,\frac{\sigma^2}{n})$, then we know that $\frac{n_i\bar{U}_i}{\sigma^2} \sim \chi^2_{n_i-1}$. So, $\bar{\bar{U}}=\frac{\sum n_iU_i/\sigma^2}{\sum n_i / \sigma^2}$. Is this the direction that I should be pursuing? $\endgroup$
    – Jen Snow
    May 4, 2020 at 23:28
  • $\begingroup$ @BruceET I also see that $s^2_k=n_1(\bar{U}_1-\bar{\bar{U}})^2+...+n_k(\bar{U}_k-\bar{\bar{U}})^2=\frac{(\bar{U}_1-...-\bar{U}_k)^2}{\frac{1}{n_1}+...+\frac{1}{n_k}}$ , where $\bar{U}_1-...-\bar{U}_k \sim N(0,\sigma^2(\frac{1}{n_1}+...+\frac{1}{n_k}))$. I just don't see how I can tie these facts together to show that the given expression follows a $\chi^2_{k-1}$ . $\endgroup$
    – Jen Snow
    May 4, 2020 at 23:31
  • $\begingroup$ @BruceET I added some updated work to the post, but I am stuck at showing independence again. Do you have any suggestions from here aside from finding the joint of $(U_1,...,U_{k-1})$ then using a Jacobian transformation to show that it factorizes? $\endgroup$
    – Jen Snow
    May 5, 2020 at 19:09

1 Answer 1


I assume the ANOVA model is

$$Y_{ij}=\theta_i+\varepsilon_{ij}\quad,\small\,i=1,2,\ldots,k\,;\,j=1,2,\ldots,n_i$$ where $\varepsilon_{ij}$'s are i.i.d $N(0,\sigma^2)$ for all $i,j$. In other words, $Y_{ij}\sim N(\theta_i,\sigma^2)$ independently $\forall\, i,j$.

Mean of the $i$th group is $$\overline {Y_{i\cdot}}=\frac1{n_i}\sum\limits_{j=1}^{n_i}Y_{ij}\quad,\, i=1,\ldots,k$$

The grand mean is then $$\overline Y=\frac{\sum_{i=1}^k n_i\overline {Y_{i\cdot}}}{\sum_{i=1}^k n_i}$$

You also defined $$\overline\theta=\frac{\sum_{i=1}^k n_i \theta_i}{\sum_{i=1}^k n_i}$$

Now $\overline {Y_{i\cdot}}\sim N\left(\theta_i,\frac{\sigma^2}{n_i}\right)$ independently for each $i$, so that

$$X_i=\overline {Y_{i\cdot}}-\theta_i\stackrel{\text{ ind.}}\sim N\left(0,\frac{\sigma^2}{n_i}\right)\quad,\,i=1,\ldots,k$$

We also have the weighted average

$$\overline X_w=\overline Y-\overline\theta=\frac{\sum_{i=1}^k n_i(\overline {Y_{i\cdot}}-\theta_i)/\sigma^2}{\sum_{i=1}^k n_i/\sigma^2}=\frac{\sum_{i=1}^k w_i X_i}{\sum_{i=1}^k w_i}\,,$$

where $w_i=\frac{n_i}{\sigma^2}$ are the weights.

As you said, the problem boils down to finding the distribution of the weighted sum of squares

$$S^2=\sum_{i=1}^k \frac{n_i}{\sigma^2}\left\{(\overline {Y_{i\cdot}}-\theta_i)-(\overline Y-\overline\theta)\right\}^2=\sum_{i=1}^k w_i(X_i-\overline X_w)^2$$

Using general facts about distributions of quadratic forms (like some form of Cochran's theorem) it can be shown that $S^2\sim \chi^2_{k-1}$, but for a more instructive derivation using orthogonal transformations you can refer to this post on Math.SE. The independence of $\overline X_w$ and $S^2$ can also be shown this way.

  • $\begingroup$ Thank you so much for providing this answer! This makes so much sense. I appreciate your detailed breakdown. $\endgroup$
    – Jen Snow
    Jun 18, 2020 at 18:30
  • $\begingroup$ This can also be answered using this theorem. $\endgroup$ Jul 11, 2020 at 14:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.