# Finding Constant for Unbiased Estimation of Standard Deviation in Normal Distribution

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?

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.