The statistical model for a randomised complete blocking design (RCBD) is $Y_{ij} = \mu + \alpha_i + \beta_j + \epsilon_{ij}$, where

$\mu$ is the overall mean, $\alpha_i$ is the effect of the $i$th treatment, $\beta_j$ is the effect of the $j$th block, and $\epsilon_{ij}$ is the Normal random error term $\epsilon_{ij} \sim N(0, \sigma^2)$ with mean 0 and variance $\sigma^2$.

I'm trying to find the expected value of $Y_{ij} = \mu + \alpha_i + \beta_j + \epsilon_{ij}$.

My understanding is that this represents the observed value of $Y_{ij}$, where the error term $\epsilon_{ij}$ represents the difference between the observed value and the expected value (that is why it is referred to as the error)? Is this correct?

And so, if I'm understanding this correctly, would the expected value just be $Y_{ij} = \mu + \alpha_i + \beta_j$?

I would greatly appreciate it if people could please take the time to clarify this. I think my explanation isn't totally clear, even to myself, so I would really appreciate it if someone who understands this better could please take the time to explain it.

  • $\begingroup$ Your understanding is correct, we write $E(Y_{ij}) = \mu+\alpha_i+\beta_j$. Note also that $E(\epsilon_{ij})=0$ as you mentioned. $\epsilon_{ij}$ models the noise around your mean but has zero mean itself. $\endgroup$ – Knarpie Apr 23 '18 at 7:39

Your intuitive understanding is correct that $Y_{ij}$ is the observed value and $\epsilon_{ij}$ is the error, so the expected value of $Y_{ij}$ would be $\mu + \alpha_i + \beta_j$.

Expanding on Knarpie's comment, to show this mathematically, we will use the properties of expectation. Namely, so a random variable $X$ and any constant $a$,

$$E(a + X) = a + E(X)\,. $$

Thus, \begin{align*} E(Y_{ij}) & = E(\mu + \alpha_i + \beta_j + \epsilon_{ij})\\ & = \mu + \alpha_i + \beta_j + E(\epsilon_{ij})\\ & = \mu + \alpha_i + \beta_j + 0\\ & = \mu + \alpha_i + \beta_j\,. \end{align*}


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