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This question appears in my textbook in a section on unbiased estimators. I know how to solve it; I'm not asking for help with the solution but for deeper understanding of the question:

Consider a random sample $X_{1},...,X_{n}$ from the pdf $f(x; \theta)=.5(1+\theta x)\quad -1\leq x \le 1$ where $-1\leq \theta \le 1$ (this distribution arises in particle physics). Show that $\hat{\theta}=3\overline{X}$ is an unbiased estimator of $\theta$. [Hint: First determine $\mu$ = E($X$) = E($\overline{X}$).]

I thought that for a function to be a PDF, the area under the curve for the specified interval has to be equal to 1. In this case, the area under f(x, $\theta$) for the interval $-1\leq x \leq 1$ is only equal to 1 if $\theta$ = 2, which isn't even in the interval of possible $\theta$s.

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  • $\begingroup$ You have the correct understanding of a pdf. Either you are integrating incorrectly, or you wrote down the density incorrectly. $\endgroup$ – Taylor Apr 2 '17 at 18:30
  • $\begingroup$ This question requires the self-study tag. This cannot be a density for all values of $\theta$ in the interval [-1, 1]. $\endgroup$ – Michael Chernick Apr 2 '17 at 18:35
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    $\begingroup$ If you were to graph $(x, f(x,\theta))$ it would immediately become clear that the areas does not depend on $\theta$, that the area is indeed $1$, and that $f$ can be a PDF only when $-1\le\theta\le 1$ (for otherwise it would attain negative values for some $x\in [-1,1]$). (@Michael: Because what you wrote seems wrong, could you enlighten me by exhibiting a value of $\theta$ in the interval $[-1,1]$ for which $f(x;\theta)$ is not a density?) $\endgroup$ – whuber Apr 2 '17 at 19:35
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I thought that for a function to be a pdf, the area under the curve for the specified interval has to be equal to 1.

That's correct.

In this case, the area under $f(x, θ)$ for the interval is only equal to 1 if $θ = 2$

Not so. $\int_{-1}^1 .5(1 + θx) \;dx = 1$ for any $θ$. For,

$$\int_{-1}^1 .5(1 + θx) \;dx = .5 \int_{-1}^1 1 \;dx + .5θ \int_{-1}^1 x \;dx = .5 \cdot 2 + .5θ \cdot 0 = 1.$$

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    $\begingroup$ Indeed, I made a mistake with the integration. Many thanks. $\endgroup$ – Jon Apr 2 '17 at 21:10

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