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


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


$X_1 \sim \mathcal{N}(\mu_1,\sigma_1^2)$ 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,

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


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

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

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


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


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

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

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

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

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