Let $\{X_1, X_2, \ldots, X_n\}$ be a random sample from a $N(0, \theta^2)$ distribution. We want to estimate the standard deviation $\theta$. Find the constant $c$ so that $\hat{\theta} = c \sum_{i=1}^{n} |X_i|$ is an unbiased estimator of $\theta$ and determine its efficiency.

My idea: Solution:

To find the constant $c$ such that $\hat{\theta} = c \sum_{i=1}^{n} |X_i|$ is an unbiased estimator of $\theta$, we use the definition of an unbiased estimator:

$ E(\hat{\theta}) = \theta $

Since $X_i$ follows a $N(0, \theta^2)$ distribution, the absolute value $|X_i|$ follows a half-normal distribution with scale parameter $\sqrt{\frac{2}{\pi}}\theta$. The probability density function (PDF) of a half-normal distribution is given by:

$f(x) = \sqrt{\frac{2}{\pi}}\frac{1}{\theta} e^{-\frac{x^2}{2\theta^2}} $

Now, let's compute the expected value:

$ E(\hat{\theta}) = c \sum_{i=1}^{n} E(|X_i|) $

$ = c \sum_{i=1}^{n} \int_{0}^{\infty} x \sqrt{\frac{2}{\pi}}\frac{1}{\theta} e^{-\frac{x^2}{2\theta^2}} \, dx $

Is my solution til now correct? how can I continue it?


1 Answer 1


Your method is absolutely correct up to this point. What's left to do is to solve the integral you've obtained and solve the equation $c\sum_{i=1}^n \int _0^{\infty} \dots dx_i$.

To do so, notice the integral contains a function of $x$: $e^{-\frac{x^2}{2\theta^2}}$ and it's inner derivative adjusted for a multiplication constant: $\frac{\partial}{\partial x}(-\frac{x^2}{2\theta^2}) = \frac{-x}{\theta^2}$.

Therefore, we can integrate using the following trick:

$$\int _0^{\infty} x \sqrt{\frac{2}{\pi}} \frac{1}{\theta} e^{-\frac{x^2}{2\theta^2}} dx = -\theta \sqrt{\frac{2}{\pi}} \int_{x = 0}^\infty - \frac{x}{\theta^2}e^{-\frac{x^2}{2\theta^2}} dx $$

$$-\theta \sqrt{\frac{2}{\pi}} \int_{x = 0}^\infty - \frac{x}{\theta^2}e^{-\frac{x^2}{2\theta^2}} dx = -\theta \sqrt{\frac{2}{\pi}} (e^{-\frac{x^2}{2\theta^2}} |_{x=0}^\infty) = \theta \sqrt{\frac{2}{\pi}} $$

Putting this expression back in your equation, we finally obtain

$$ E(\hat{\theta}) = cn\theta \sqrt{\frac{2}{\pi}} $$

Using the definition of an unbiased estimator and solving for $c$ will give you the answer.


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