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Suppose $Y$ is discrete and only takes on non-negative integers and that the conditional distribution of $Y$ given $X=x$ is Poisson, that is, $$P(Y=y|X=x) = \frac{\exp(-x'\beta) (x'\beta)^y}{y!}$$ where $y = 0, 1, 2, \cdots$. First compute $E(Y|X=x)$ and $Var(Y|X=x)$, does this justify a linear regression model of the form $y = x'\beta + e$?

I have calculated $E(Y|X=x) = Var(Y|X=x) = x'\beta$ by the properties of a Poisson distribution. I am unsure how to answer the last part of the question related to the linear regression model. Could anyone shed some ideas?

elbarto
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    Notice that "$\beta$" has different meanings in the two models you are asked to compare. Maybe it would be clearer if the question asked you to evaluate a regression model of the form $y=x^\prime\gamma+e$. Given that the Poisson model applies, can you work out the distribution the random variable $e$ must have, conditional on $x$? If not the full distribution (that's hard), at least its expectation and variance? – whuber Oct 19 '17 at 14:41
  • OK, so here's what I did: 1) The conditional expectation $E(e|x) = E(y-x'\gamma|x) = E(y|x) - E(x'\gamma|x) = x'\gamma - x'\gamma = 0$. 2) The conditional variance $V(e|x) = E(e^2|x) - E(e|x)^2 = E(y^2-2yx'\gamma + (x'\gamma)^2|x) = E(y^2|x) - 2x'\gamma E(y|x)+ (x'\gamma)^2 = E(y^2|x)-(E(y|x))^2 = V(y|x) = x'\gamma$.

    Is this correct? How can I relate it back to the linear regression?

     – elbarto Oct 19 '17 at 22:56
  • I don't think it's right, because you don't seem to have distinguished the two models from each other. The expectation $E(e\mid x)$ will generally not be zero because in the correct model (the first) the regression is exponential. When you attempt to fit a linear regression to this, the errors cannot average out to zero at any more than two $x$ values. – whuber Oct 19 '17 at 23:05
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    Oh I see what you mean now. $E(y|x) = x'\beta$ while $E(x'\gamma|x) = x'\gamma$, so we have $E(e|x) = x'\beta - x'\gamma$ and this equals $0$ if and only if $\beta = \gamma$, which doesn't have to be always be true. So a linear regression model here is not justified because the errors do not always average out to zero. Is this correct? – elbarto Oct 19 '17 at 23:09
  • That is how I interpret the question. There might be other ways, but I think it gets to the heart of the matter. – whuber Oct 20 '17 at 13:08
  • Related: https://stats.stackexchange.com/questions/142338/goodness-of-fit-and-which-model-to-choose-linear-regression-or-poisson/142353#142353 – kjetil b halvorsen Jun 26 '21 at 00:22

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