Deriving variance component of 1-Way ANOVA?

My goal is to drive $$E(MSA)$$ for this design(the 1-way ANOVA); $$y_{ij}=\mu +\alpha_i + \epsilon_{ij}$$for i=1,2,...,a (that is, there are a groups)and j=1,2,...,n. $$\epsilon_{ij}$$ and $$\alpha_i$$ are random as follows;$$E(\alpha_i)=0, E(\epsilon_{ij})=0$$ $$var(\alpha_i)=\sigma_{\alpha}^2, var(\epsilon_{ij})=\sigma_{\epsilon}^2$$and all the covariances are zero.

The answer shoud be $$E(MSA)=n\sigma_{\alpha}^2+\sigma_{\epsilon}^2$$.Could I let mathematica derive the answer? I have tried inputting as below naively, but I couldn't get the answer.

    Subscript[y, i] = (*definition for average value yi for each group i*)
Sum[Subscript[y, i, j],
{j, 1, n}]/n

Subscript[y] =  (*definition for average of all yij*)
Sum[Subscript[y, i, j],
{i, 1, a}, {j, 1, n}]/
(a*n)

Expectation[(n/(a - 1))*
Sum[(Subscript[y, i] -
Subscript[y])^2,
{i, 1, a}],
{Distributed[Subscript[y,
i], NormalDistribution[
0, Subscript[\[Sigma], \[Alpha]]]],
Distributed[Subscript[y],
NormalDistribution[0,
Subscript[\[Sigma], \[Epsilon]]]]}]


• Please include all other definitions in your code, e.g. $y_i$ etc. Also see: How to copy code from Mathematica so it looks good on this site. Commented Jun 6, 2023 at 13:50
• Either your definition of $MSA$ or $E(MSA)$ is wrong. As evidence for that consider $a=1$. Then Sum[(Subscript[y, i] - Subscript[y])^2, {i, 1, a}] is identically 0 and $n \sigma_\alpha^2+\sigma^2$ can't be zero. Also, you should use indexed variables rather than Subscript. See mathematica.stackexchange.com/questions/245319/….
– JimB
Commented Jun 6, 2023 at 17:13
• I suspect you want to multiply Sum[(Subscript[y, i] - Subscript[y])^2, {i, 1, a}] by n/(a - 1).
– JimB
Commented Jun 6, 2023 at 17:29

Here are three approaches that work for specified values of a and n:

For the first two approaches perform the following:

a = 6;
n = 7;
y = Table[μ + α[i] + ϵ[i, j], {j, 1, n}, {i, 1, a}];
overallMean = Total[Total[y]]/(n a);
groupMean = Total[y]/n;
msa = (n/(a - 1)) Sum[(groupMean[[i]] - overallMean)^2, {i, 1, a}] // Expand;


Approach 1: Using TransformedDistribution

dist = TransformedDistribution[msa, Flatten[{
Table[α[i] \[Distributed] NormalDistribution[0, σα], {i, 1, a}],
Table[ϵ[i, j] \[Distributed] NormalDistribution[0, σ], {i, 1, a}, {j, 1, n}]}]];
Mean[dist] // Expand
(* σ^2 + 7 σα^2 *)


Approach 2: Using substitution.

Emsa = msa /. {α[i_]^2 -> σα^2, α[i1_] α[i2_] -> 0, ϵ[i_, j_]^2 -> σ^2,
α[i1_] ϵ[i2_, j_] -> 0, ϵ[i1_, j1_] ϵ[i2_, j2_] -> 0}
(* σ^2 + 7 σα^2 *)


Approach 3: Knowledge of distribution of sample group means

If it is recognized that the sample group means all have a mean of $$\mu$$ and variance $$\sigma_\alpha^2+\sigma^2/n$$, then one can use the following:

msa = (n/(a - 1)) Sum[(ybar[i] - Sum[ybar[j], {j, 1, a}]/a)^2, {i, 1,  a}]
Emsa = (msa // Expand) /. ybar[i_]^2 -> σα^2 + σ^2/n + μ^2 /. ybar[i_] -> μ // Expand
(* σ^2 + n σα^2 *)


With this 3rd approach we get the general result when giving a specific value of $$a$$.

Maybe someone else knows how to do this in a more general way.

• Thanks for the accept but you might want to wait a day or two before accepting an answer because someone might give the general answer (and I'd certainly like to see that, too).
– JimB
Commented Jun 7, 2023 at 15:34