0
$\begingroup$

I am sorry for the bad heading. It was difficult to find it. What I am trying to calculate is the expectation value of (bias corrected) variance over mean squared for 2 random number drawn from the same normal distribution.

From a simple Monte-Carlo simulation I am pretty sure the solution should be:

$$<A> = \frac{\sigma ^2}{\mu ^2} $$

However, I did not find a proof online and was not able to calculate it myself. To find the expectation value one has to solve the integral

$$<A> = \frac{1}{2\pi \sigma ^2} \int \text{d}x_1\int \text d x_2 \frac{(x_1-x_2)^2}{(x_1+x_2)^2}\exp{\left(-\frac{(x_1-\mu ^2)}{2\sigma ^2}\right)}\exp{\left(-\frac{(x_2-\mu ^2)}{2\sigma ^2}\right)}$$

The beginning is pretty easy. I tried the substitution $$a=x_1-x^2\\b=x_1+x_2$$ to split the integral to $$<A> = \frac{1}{4\pi \sigma ^2} \exp(-\frac{\mu^2}{\sigma^2})\int \text{d}a~a^2 \exp{\left(-\frac{a^2}{4\sigma ^2} \right)} \int \text d b~b^{-2} exp{\left(-\frac{b^2-4b\mu}{4\sigma ^2}\right)}$$ While the first integral is not very complicated, the second is more problematic. There is a pole at $b=0$ but since, the Gaussian diverges for many points on the infinite circle, I cannot use the residue theorem. I was not able to find an antiderivative and I lack of plan C.

In order to get a nicer integral I tried to substitute $c = b+2\mu$ to get $$<A> = \frac{16 \sigma}{\sqrt{\pi}} \int \text d x~(x+2\mu)^{-2} exp{\left(-\frac{x^2}{4\sigma ^2}\right)}$$

I would be glad about every help

Edit:

I do not longer think, that my expected result is true. This should only be the limit for $|\mu|>>\sigma$

$\endgroup$
1
$\begingroup$

$X_1 \sim \mathcal{N}(\mu_1,\sigma_1^2)$ and $X_2 \sim \mathcal{N}(\mu_2, \sigma_2^2)$ and $\operatorname{Cor}(X_1,X_2) = \rho$. We will compute an approximation of,

$$ \mathbb{E}\left(\frac{X_1-X_2}{X_1+X_2}\right)^{2} $$

Let's try using delta method. Suppose $g(t_1,t_2) = \frac{t_1-t_2}{t_1+t_2}$. The first order taylor approximation of this function about $\mu_1,\mu_2$ will be,

$$ g(\mathbf{t}) \approx g(\boldsymbol{\mu}) + \sum_{i=1}^{2}(t_i-\mu_i)\frac{\partial g(\mathbf{t})}{\partial t_i}\bigg\vert_{\mathbf{t} = \boldsymbol{\mu}} $$

where,

$$ g(\boldsymbol{\mu}) = \frac{\mu_1-\mu_2}{\mu_1+\mu_2}, \ \ \ \ \frac{\partial g(\mathbf{t})}{\partial t_1} \bigg\vert_{\mathbf{t} = \boldsymbol{\mu}} = \frac{2\mu_2}{(\mu_1+\mu_2)^2}, \ \ \ \ \frac{\partial g(\mathbf{t})}{\partial t_2} \bigg\vert_{\mathbf{t} = \boldsymbol{\mu}} = \frac{-2\mu_1}{(\mu_1+\mu_2)^2} $$

Put $t_1 = X_1$ and $t_2 = X_2$,

$$ \mathbb{E}g(\mathbf{t}) \approx g(\boldsymbol{\mu}) + 0 = \frac{\mu_1-\mu_2}{\mu_1+\mu_2} $$

and,

$$ \begin{align} \operatorname{Var}g(\mathbf{t}) &= \mathbb{E}(g(\mathbf{t}) - g(\boldsymbol{\mu}))^2 \approx \sum_{i=1}^{2}\sum_{j=1}^{2}\operatorname{Cov}(t_i-\mu_i, t_j-\mu_j)\frac{\partial g(\mathbf{t})}{\partial t_i}\bigg\vert_{\mathbf{t} = \boldsymbol{\mu}}\frac{\partial g(\mathbf{t})}{\partial t_j}\bigg\vert_{\mathbf{t} = \boldsymbol{\mu}} \\ &= \sigma_1^2\left(\frac{\partial g(\mathbf{t})}{\partial t_1}\right)^2\bigg\vert_{\mathbf{t} = \boldsymbol{\mu}} + \sigma_2^2\left(\frac{\partial g(\mathbf{t})}{\partial t_2}\right)^2\bigg\vert_{\mathbf{t} = \boldsymbol{\mu}} + 2\rho\sigma_1\sigma_2\frac{\partial g(\mathbf{t})}{\partial t_1}\bigg\vert_{\mathbf{t} = \boldsymbol{\mu}}\frac{\partial g(\mathbf{t})}{\partial t_2}\bigg\vert_{\mathbf{t} = \boldsymbol{\mu}} \\ &= \frac{4(\mu_1^2\sigma_1^2 + \mu_2^2\sigma_2^2 - 2\rho\sigma_1\sigma_2\mu_1\mu_2)}{(\mu_1+\mu_2)^4} \end{align} $$

Finally,

$$ \mathbb{E}g(\mathbf{t})^2 = \operatorname{Var}g(\mathbf{t}) + (\mathbb{E}g(\mathbf{t}))^2 \approx \frac{4(\mu_1^2\sigma_1^2 + \mu_2^2\sigma_2^2 - 2\rho\sigma_1\sigma_2\mu_1\mu_2) + (\mu_1-\mu_2)^2(\mu_1+\mu_2)^2}{(\mu_1+\mu_2)^4} $$

Now, put $\mu_1 = \mu_2 = \mu$, $\sigma_1 = \sigma_2 = \sigma$ and $\rho = 0$, to get,

$$ \mathbb{E}\left(\frac{X_1-X_2}{X_1+X_2}\right)^{2} \approx \frac{\sigma^2}{2\mu^2} $$

$\endgroup$
  • $\begingroup$ Thank you, this is a very interesting approach. I still have to think about the Taylor expansion. Maybe my expected solution is simply not true and it is only valid for $\mu >> \sigma$ $\endgroup$ – Glostas Aug 23 '17 at 13:25
  • $\begingroup$ I think that the error term of above approximation will lower if $|\mu| >> 0$ and $\sigma$ has a small positive value. I think you must have observed this in your monte-carlo approximation that, when $\mu$ is closer to $0$ or when $\sigma$ is very high, the expected value explodes. May be the integral diverges for certain set of values of $\mu$ and $\sigma$. This requires more analysis. Will try to do it if time permits. $\endgroup$ – Dhruv Kohli - expiTTp1z0 Aug 23 '17 at 13:35
  • $\begingroup$ well, you are right. I do see that. I only tested values close to the expected data. This was not very smart;) $\endgroup$ – Glostas Aug 23 '17 at 14:04
  • $\begingroup$ I could adapt your solution for every number of picks (not only 2, like in this question). It is really what I want $\endgroup$ – Glostas Aug 24 '17 at 11:34
  • $\begingroup$ Happy to help. By the way, to read more on this, check out The Delta Method in chapter on Properties of a Random Sample, Statistical Inference, Second Edition, G.Casella and R.L.Berger. $\endgroup$ – Dhruv Kohli - expiTTp1z0 Aug 24 '17 at 12:29
1
$\begingroup$

As you have noticed for a Gaussian random variable $X \sim \mathcal{N}(\mu, \sigma^2)$ the first and higher order moments of the variable $1/X$ do not exist. In your choice of substitution you actually found that, were it to exist $$ \mathbb{E} \left[ \left(\frac{(X_1 - X_2)}{(X_1 + X_2)} \right)^2 \right] = \mathbb{E}\left[ (X_1-X_2)^2 \right]\mathbb{E}\left[ \frac{1}{(X_1 + X_2)^2 }\right] $$ Well since $X_1$ and $X_2$ are independent draws from the common distribution $\mathcal{N}(\mu_x, \sigma^2_x)$ then you have that $U = X_1 - X_2$ and $V = X_1 + X_2$ are independent Gaussian random variables, see this question.

It further follows that $U^2$ and $V^2$ are independent non central $\chi^2$ random variables. The distribution of the ratio of $U^2$ and $V^2$ would be given by the $F$-distribution, or rather the generalisation to the non-central case, but again note that for the moments of the $F$-distribution to exist you need the degrees of freedom for the reciprocal variable to be greater than $2$, that is to say the expectation of the quantity you are trying to find is undefined when only looking at a pair of random variables.

$\endgroup$
  • $\begingroup$ Thank you, I do not think, that your formula matches the finding of my substitution. The first term of your RHS matches the $\text d a$ integration. However, the second term does not. Since there is the offset of $2 \mu$ in the integral, it does not match the shown expectation value. Therefore, the statement that "the expectation of the quantity" I am "trying to find is undefined" is only valid for $\mu = 0$. $\endgroup$ – Glostas Aug 23 '17 at 14:30
  • $\begingroup$ $b = X_1 + X_2$ is a mean $2\mu$ Gaussian random variable, just pull the constant term $\exp(-\mu^2/\sigma^2)$ through into the square of the last exponential and it should match up? $\endgroup$ – Nadiels Aug 23 '17 at 14:45
  • $\begingroup$ You mean the integrant shall be $\frac{(\exp{-(b-2\mu)^2)}/4\sigma ^2}{b^2}$ and since b has a mean of $2\mu$ and a deviation of $\sqrt{2}\sigma$ this is proportional to the calculation of an expectation value? $\endgroup$ – Glostas Aug 23 '17 at 15:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.