Suppose that $Y = a + b K + X$ with $0 < X$ and $0 < K$ where $X, Y$ and $K$ are random variables. What are then the expectations of the intercept and slope in the case of a linear regression of $Y$ on $K$?

  • 2
    $\begingroup$ it depends on the correlation between $X$ and $K$. $\endgroup$
    – AdamO
    Sep 20, 2019 at 13:42
  • 1
    $\begingroup$ You need to correct your question to indicate that you are interested in the expectations of the estimated intercept and slope. As it stands, the question is ambiguous. $\endgroup$ Sep 20, 2019 at 14:38
  • $\begingroup$ In conducting linear regression one generally is concerned about the conditional responses and therefore the expectations conditional on $K.$ Those can be computed quite generally, but the full expectations--accounting for the distribution of $K$--cannot be generally computed: there's no nice formula for them. Which are you asking for? $\endgroup$
    – whuber
    Sep 20, 2019 at 19:45
  • $\begingroup$ To AdamO X and K are independent. $\endgroup$ Sep 21, 2019 at 8:37
  • $\begingroup$ To Isabella Ghement I am interested in the expectations of the estimated intercept and slope. $\endgroup$ Sep 21, 2019 at 8:39

1 Answer 1


Just to be sure we share a common understanding of the question, I suppose you mean you have independent data $(X_i, K_i, y_i)$ where $(K_i,X_i)$ are $n$ independent and identically distributed random variables (you indicate they are positive, but this does not matter), the individual $(K_i, X_i)$ are independent (as you state in a comment), and you use ordinary least squares to estimate $a$ and $b$ in the model

$$y_i = a + b K_i + X_i.$$

Now, because you assumed $X \gt 0,$ necessarily $E[X_i] \gt 0$ (assuming these expectations are finite). This is where your question differs from the usual regression setting. The $X_i$ play the role of the "errors" in the model, but because they do not have a zero expectation, their expectation must be added to the constant $a.$

One way to see this is to write $$\varepsilon_i = X_i - E[X_i] = X_i - \mu$$ (where $\mu=E[X_i]$ is the common expectation). These random variables are now legitimate errors because they have zero expectations. Writing for the common expectation, your model is the same as

$$y_i = a + b K_i + (\mu + \varepsilon_i) = (a + \mu) + b K_i + \varepsilon_i.$$

Thus, as always in least squares regression,

the expectation of the slope estimate is $b$ and the expectation of the intercept estimate is the true intercept, here seen equal to $a + \mu.$

Unless you know $\mu,$ you cannot determine $a$ from this model: you can only estimate the sum $a+\mu.$ (It sounds like your simulations used zero-mean variables $X_i,$ contrary to the assumptions stated in the question. Indeed, the expectation of the estimate of $a$ equals $a$ if and only if $\mu=0.$)


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