I have $X_i|b_i \sim Poisson(\lambda_i)$, and $log(\lambda_i)=x_i^t\beta+\varepsilon_i$ further $\varepsilon\sim N(0,\sigma^2)$

then, how to find $E(X_i)$?

I try using the iterated expectation $E(X_i)=E[E(X_i|b_i)]$, so $E(X_i)=E(\lambda_i)$ but how to use $log(\lambda)$ to find the solution?


  • $\begingroup$ What is the $x_i^t$ in your equation for $\log(\lambda_i)?$ A constant unrelated to $X_i$? $\endgroup$ – spaceisdarkgreen Feb 20 '17 at 2:01
  • $\begingroup$ linear model $x^t_i \beta +\varepsilon$ $\endgroup$ – albert Feb 20 '17 at 2:04
  • $\begingroup$ so the $x_i$ is unrelated to the $X_i$? Why not call $X_i$ something different? $\endgroup$ – spaceisdarkgreen Feb 20 '17 at 2:06
  • $\begingroup$ Is different $x_i$ and $X_i$ $\endgroup$ – albert Feb 20 '17 at 2:07

Your iterated expectations strategy looks good.

We have $\lambda_i = Ce^{\epsilon_i}$ where $C=e^{x_i^t\beta}$ so can take the expected value $$E(\lambda_i) = CE(e^{\epsilon_i})=C\int_{-\infty}^\infty e^t \frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{t^2}{2\sigma^2}}dt $$ since $\epsilon_i\sim N(0,\sigma^2).$ The integral is a Gaussian integral and can be done by completing the square.

  • $\begingroup$ Why does the expected value have that form? That is, why f (x)=N(0,1) $\endgroup$ – albert Feb 20 '17 at 2:17
  • 1
    $\begingroup$ @albert I'm using the fact that $\epsilon_i \sim N(0,\sigma^2).$ Since $\lambda_i =e^{x_i^t\beta}e^{\epsilon_i},$ we have$E(\lambda_i) = e^{x_i^t\beta}E(e^{\epsilon_i}).$ So you need to compute $E(e^Z)$ where $Z\sim N(0,\sigma^2)$ which is $\int e^z f_Z(z)dz$ where $f_Z(z)$ is the PDF for a $N(0,\sigma^2)$. $\endgroup$ – spaceisdarkgreen Feb 20 '17 at 2:28

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