I am working through a problem where I need to prove that $E\left[\hat{\theta}_{\ln}\right]\leq\theta-1$, where $\hat{\theta}_{\ln}=\frac{1}{n}\sum_{i=1}^{n}\ln(X_{i})$.

I start from $E\left[\ln\left(X_{i}\right)\right]\leq E\left[X_{i}\right]-1 .$

Would it be wrong simply to write

$E\left[\hat{\theta}_{\ln}\right] =E\left[\frac{1}{n}\sum_{i=1}^{n}\ln(X_{i})\right] =\frac{1}{n}\sum_{i=1}^{n}E\left[\ln(X_{i})\right] =E\left[\ln(X_{i})\right]$ and then use $E\left[X_{i}\right]=\theta $, which is previously assumed?

  • $\begingroup$ About which of these steps are you uncertain? $\endgroup$ – whuber Oct 26 '15 at 17:59
  • $\begingroup$ The step where $E\left[\hat{\theta_{\ln}}\right]=E\left[\ln(X_{i})\right]$ $\endgroup$ – avriis Oct 26 '15 at 18:08

If your concern is the last step, then you might want to read a little bit more on the expectation operator. Its basic property is that it is a linear operator and so the expectation of the sum becomes the sum of the expectations. Assuming your variables constitute a random sample, i.e. they are iid, their means do not differ and so you are taking $n$ times the same mean. Thus the multiple $\frac{1}{n}$ cancels and you are left with $E\left[ \log(X) \right]$.

I believe the trickiest part of this exercise was to understand why $E\left[\log\left(X\right)\right]\leq E\left[X\right]-1$ but no problems there, right?

  • $\begingroup$ Thanks! Yes, $E\left[\log\left(X\right)\right]\leq E\left[X\right]-1$ was indeed more tricky, but I figured that out. I just wasn't certain whether the usual properties of the expectation operator applied with a logarithmic function. $\endgroup$ – avriis Oct 27 '15 at 8:51

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