# Could you prove the theorem? The theorem 1 is the result in Emil Bjornson's paper (PILOT-BASED BAYESIAN CHANNEL NORM ESTIMATION IN RAYLEIGH FADING MULTI-ANTENNA SYSTEMS).

I want to know the proof omitted. Please, help me.

I don't think this question is a better fit on math.SE as Xi'an avers (actually dsp.SE might be a better fit where a lot of digital communications systems questions seem to end up) nor do I agree that it is off-topic here. I don't have a complete answer for the OP but nonetheless, here goes.

Let $$[Y_1, Y_2] = [H_1, H_2] + [N_1, N_2]$$ where the $$H_i$$ are i.i.d. $$N(0,\sigma_h^2)$$ and the $$N_i$$ are i.i.d. $$N(0,\sigma_n^2)$$, and the $$H$$'s and $$N$$'s are also mutually independent (the OP does not state this but I think it is implicit in the model). Then, it is easily seen that the $$Y_i$$ are i.i.d. $$N(0,\sigma_h^2 + \sigma_n^2)$$ random variables.

As background, in a digital communications systems context, $$[H_1, H_2]$$ is the measurement of the two components of a Rayleigh faded signal $$\big($$the signal amplitude $$\sqrt{H_1^2+H_2^2}$$ is a Rayleigh random variable$$\big)$$ received in Gaussian noise (which results in the actual measurement $$[Y_1, Y_2] = [H_1, H_2] + [N_1, N_2],$$ where $$[N_1,N_2]$$ is the Gaussian noise in the measurement of the two components). The question thus is to find the conditional distribution of the signal energy $$\mathscr{E}_h = H_1^2+H_2^2$$ given the value of the received energy $$\mathscr{E}_y = Y_1^2+Y_2^2$$. Note that both $$\mathscr{E}_h$$ and $$\mathscr{E}_y$$ are exponential random variables.

Now, in the digital communications literature, the distribution of $$\sqrt{X^2+Y^2}$$ where $$X$$ and $$Y$$ are independent Gaussian random variables with nonzero means $$\alpha_1$$ and $$\alpha_2$$ and common variance $$\sigma^2$$ is called the Ricean distribution (after S.O. Rice, "Mathematical Analysis of Random Noise," Bell System Technical Journal, 1944, 1945) whose density is $$f_{\sqrt{X^2+Y^2}}(u) = \frac{u}{\sigma^2}\exp\left(-\frac{u^2+\alpha_1^2+\alpha_2^2}{2\sigma^2}\right)I_0\left(\frac{u\sqrt{\alpha_1^2+\alpha_2^2}}{\sigma^2}\right), ~u \geq 0.\tag{1}$$ It follows that the density of $$X^2+Y^2$$ is given by \begin{align}f_{X^2+Y^2}(v) &= \frac{1}{\sqrt{v}}f_{\sqrt{X^2+Y^2}}(\sqrt{v})\\ &= \frac{1}{\sigma^2}\exp\left(-\frac{v+\alpha_1^2+\alpha_2^2}{2\sigma^2}\right)I_0\left(\frac{\sqrt{v(\alpha_1^2+\alpha_2^2)}}{\sigma^2}\right), ~v \geq 0.\tag{2}\end{align}

We use the results in $$(1)$$ and $$(2)$$ to attain the desired result as follows. For $$i=1,2$$, $$Y_i$$ and $$H_i$$ are zero-mean jointly normal random variables with respective variances $$\sigma_h^2 + \sigma_n^2$$ and $$\sigma_h^2$$ and covariance $$\sigma_h^2$$. Hence, conditioned on $$Y_i=\alpha_i$$, the conditional distribution of $$H_i$$ is normal with mean $$\frac{\sigma_h^2}{\sigma_h^2 + \sigma_n^2}\alpha_i$$ and variance $$\frac{\sigma_h^2\sigma_n^2}{\sigma_h^2+\sigma_n^2}$$. Note that the variance does not depend on the value of $$\alpha_i$$. Furthermore, since $$[Y_1,H_1]$$ is independent or $$[Y_1.H_2]$$, then conditioned on $$[Y_1,Y_2] = [\alpha_1, \alpha_2]$$, the random variables $$H_1$$ and $$H_2$$ are independent normal random variables with means as specified above and common variance $$\frac{\sigma_h^2\sigma_n^2}{\sigma_h^2+\sigma_n^2}$$. Thus, we can use $$(2)$$ to write down that for $$v \geq 0$$, $$f_{\mathscr{E}_h \mid [Y_1,Y_2] = [\alpha_1, \alpha_2]}(v\mid\alpha_1, \alpha_2) \\= \frac{\sigma_h^2+\sigma_n^2}{\sigma_h^2\sigma_n^2} \exp\left(-\frac{\sigma_h^2+\sigma_n^2}{2\sigma_h^2\sigma_n^2}v\right) \exp\left(-\frac{\sigma_h^2}{2\sigma_n^2(\sigma_h^2+\sigma_n^2)}(\alpha_1^2+\alpha_2^2)\right) I_0\left(\frac{\sqrt{v(\alpha_1^2+\alpha_2^2)}}{\sigma_n^2}\right).~~(3)$$ But what if instead of knowledge of the values $$\alpha_1$$ and $$\alpha_2$$ of $$Y_1$$ and $$Y_2$$ separately, we only know the value of $$Y_1^2+Y_2^2$$, that is, the value $$E_y$$ of the random variably $$\mathscr{E}_y$$? Well, as $$(3)$$ shows, we don't really need the values of $$\alpha_1$$ and $$\alpha_2$$, and we have that $$f_{\mathscr{E}_h \mid \mathscr{E}_y = E_y}(v\mid E_y) \\= \frac{\sigma_h^2+\sigma_n^2}{\sigma_h^2\sigma_n^2} \exp\left(-\frac{\sigma_h^2+\sigma_n^2}{2\sigma_h^2\sigma_n^2}v\right) \exp\left(-\frac{\sigma_h^2}{2\sigma_n^2(\sigma_h^2+\sigma_n^2)}E_y\right) I_0\left(\frac{\sqrt{vE_y}}{\sigma_n^2}\right), v \geq 0.~~(4)$$ Replacing $$\sigma_h^2$$ and $$\sigma_n^2$$ by $$\frac{\lambda}{2}$$ and $$\frac{\mu}{2}$$ respectively, we get the pdf $$f_{\mathscr{E}_h \mid \mathscr{E}_y = E_y}(v\mid E_y) = \frac{\lambda+\mu}{\lambda\mu} \exp\left(-\frac{\lambda+\mu}{\lambda\mu}v\right) \exp\left(-\frac{\lambda}{\mu(\lambda+\mu)}E_y\right) I_0\left(\frac{2}{\mu}\sqrt{vE_y}\right), v \geq 0.\tag{5}$$

exhibited in the paper that the OP is reading.

• Thank you so much.I knew you could look at the problem from the perspective you mentioned. – W W W Jan 18 '19 at 5:20
• how to derive the expectation and variance? how to deal with Bessel function, deriving the expectation and variance? – W W W Jan 28 '19 at 12:50

A brief description of the derivation below:

• find the distribution of $$H$$ conditional on $$y$$ (using Bayes rule)
• transform the density to be expressed in coordinates $$\varrho_h$$ and $$\theta$$
• then integrate over the region where $$\varrho_h$$ is constant, ie a circle (resulting into that Bessel function).

### Distribution of $$H$$ conditional on $$y$$

The individual distributions are:

$$\begin{array}{rcl} f_Y(y) & = & \frac{1}{\pi (\lambda+\mu)} e^{-(\lambda + \mu)^{-1}\vert y \vert^2} \\ f_H(h) & = & \frac{1}{\pi \lambda} e^{-\lambda^{-1}\vert h \vert^2} \\ f_N(n) & = & \frac{1}{\pi \mu} e^{-\mu^{-1}\vert n \vert^2} \end{array}$$

From which you can derive a conditional distribution (the last step uses the law of cosines):

$$\begin{array}{rcl} f_{H \vert Y}(h \vert y) &=& \frac{f_{Y \vert H}(y \vert h) f_H(h) }{f_Y(y)} \\ & = & \frac{f_{N}(y - h) f_H(h) }{f_Y(y)} \\ & = & \pi^{-1} \frac{\lambda + \mu}{\lambda \mu} \, e^{-\lambda^{-1}\vert h \vert^2} \, e^{(\lambda + \mu)^{-1}\vert y \vert^2} \, e^{-\mu^{-1}\vert y-h \vert^2} \\ & = & \pi^{-1} \frac{\lambda + \mu}{\lambda \mu} \, e^{-\frac{\lambda+\mu}{\lambda \mu}\vert h \vert^2} \, e^{-\frac{\lambda}{\mu (\lambda+\mu)}\vert y \vert^2} \, e^{\frac{2}{\mu} \vert y \vert \vert h \vert cos(\theta)}\\ \end{array}$$

where $$\theta$$ is the angle between $$h$$ and $$y$$.

### Coordinate transformation

Note: the density above is for the complex number expressed by Cartesian coordinates ($$dx$$ and $$dy$$), we will later use some sort of polar coordinates ($$d \theta$$ and $$d(r^2)$$ or $$d(\varrho_h)$$). The density will transform due to the transformation of the differential element $$dA = dxdy = r drd\theta = 0.5 d(r^2)d\theta = 0.5 d\varrho_h d\theta$$

In polar coordinates $$\theta$$ and $$r$$ you get:

$$\begin{array}{rcl} f_{H \vert Y}(h \vert y) &=& r \pi^{-1} \frac{\lambda + \mu}{\lambda \mu} \, e^{-\frac{\lambda+\mu}{\lambda \mu}r^2} \, e^{-\frac{\lambda}{\mu (\lambda+\mu)}\varrho_y} \, e^{\frac{2}{\mu} \sqrt{r^2 \varrho_y}cos(\theta)}\\ \end{array}$$

In coordinates $$\theta$$ and $$\varrho_h = r^2$$ you get:

$$\begin{array}{rcl} f_{H \vert Y}(h \vert y) &=& 0.5 \pi^{-1} \frac{\lambda + \mu}{\lambda \mu} \, e^{-\frac{\lambda+\mu}{\lambda \mu}\varrho_h} \, e^{-\frac{\lambda}{\mu (\lambda+\mu)}\varrho_y} \, e^{\frac{2}{\mu} \sqrt{\varrho_h \varrho_y} cos(\theta) }\\ \end{array}$$

### Integration to get from '$$H$$ conditional on $$y$$' to '$$\varrho_H$$ conditional on $$y$$'

If you integrate this for $$0 \leq \theta< 2\pi$$ you get$$^\dagger$$:

$$\begin{array}{rcl} f_{\varrho_H = \varrho_h \vert \varrho_Y}(\varrho_h \vert \varrho_y) &=& f_{\varrho_H = \varrho_h \vert Y}(\varrho_h \vert y) \\ &=& \int_0^{2\pi} f_{H \vert Y}(h \vert y) d\theta \\ &=& \frac{\lambda + \mu}{\lambda \mu} \, e^{-\frac{\lambda+\mu}{\lambda \mu}\varrho_h} \, e^{-\frac{\lambda}{\mu (\lambda+\mu)}\varrho_y} \, I_0\left(\frac{2}{\mu} \sqrt{\varrho_h \varrho_y} \right)\\ \end{array}$$

using the property $$\int_0^{2\pi} e^{x \cos(\theta)}d\theta = 2\pi I_0(x)$$ and noting that $$f_{\varrho_H = \varrho_h \vert \varrho_Y}(\varrho_h \vert \varrho_y) = f_{\varrho_H = \varrho_h \vert Y}(\varrho_h \vert y)$$ for any $$y$$ due to the spherical symmetry.

### Finding raw moments

In order to find the $$k$$-th raw moment you would have to integrate

$$\int \varrho_h^k f(\varrho_h) d \varrho_h$$

or replacing $$\varrho_h$$ by $$z^2$$ $$\int z^{2k+1} f(z^2) d z$$

which is basically (up to some constant) an integral of the form

$$\int z^{2k+1} e^{-z^2} I_0(a z) dz$$

(I have for the moment no solution for that)

$$\dagger$$ this is a bit direct. You could also first integrate over the disk (angle and radius), instead of over the circle, to get something like $$P(\varrho_H, which is the cdf, and then differentiate over $$x$$ to get the pdf.

• Thank you so much. It has helped too much. I learned a lot from you. – W W W Jan 18 '19 at 5:18
• how to derive the expectation and variance? how to deal with Bessel function, deriving the expectation and variance? – W W W Jan 28 '19 at 12:50

Cumulant generating function: The density has an exponential term in it, so finding the moment generating function is pretty simple. First we define the transformed parameter:

$$\lambda' \equiv \frac{\lambda}{1-t \lambda} \quad \quad \quad \iff \quad \quad \quad \frac{\lambda+\mu}{\lambda \mu} - t = \frac{\lambda'+\mu}{\lambda' \mu}.$$

With a bit of algebra, we then have:

\begin{equation} \begin{aligned} m(t) \equiv \mathbb{E} [ e^{t \varrho_h} | \varrho_y ] &= \int \limits_0^\infty e^{t \varrho_h} f(\varrho_h|\varrho_y) \ d\varrho_h \\[6pt] &= \int \limits_0^\infty \frac{\lambda + \mu}{\lambda \mu} e^{- \varrho_h \Big( \tfrac{\lambda + \mu}{\lambda \mu} - t \Big)} e^{- \varrho_y \Big( \tfrac{\lambda}{\mu (\lambda + \mu)} \Big)} I_0 \Big( \frac{2}{\mu} \sqrt{\varrho_h \varrho_y} \Big) \ d\varrho_h \\[6pt] &= \frac{\lambda + \mu}{\lambda + \mu - t \lambda \mu} \cdot e^{- \varrho_y \Big( \tfrac{\lambda}{\mu (\lambda + \mu)} - \tfrac{\lambda}{\mu (\lambda + \mu (1-t\lambda))} \Big)} \\[6pt] &\quad \times \int \limits_0^\infty \Big( \frac{\lambda + \mu}{\lambda \mu} - t \Big) e^{ - \varrho_h \Big( \tfrac{\lambda + \mu}{\lambda \mu} -t \Big)} e^{- \varrho_y \Big( \tfrac{\lambda}{\mu (\lambda + \mu (1-t\lambda))} \Big) } I_0 \Big( \frac{2}{\mu} \sqrt{\varrho_h \varrho_y} \Big) \ d\varrho_h \\[6pt] &= \frac{\lambda + \mu}{\lambda + \mu - t \lambda \mu} \cdot \exp \Big( \varrho_y \cdot \frac{t \lambda^2}{(\lambda + \mu) (\lambda + \mu (1-t \lambda))} \Big) \\[6pt] &\quad \times \int \limits_0^\infty \frac{\lambda' + \mu}{\lambda' \mu} e^{ - \varrho_h \Big( \tfrac{\lambda' + \mu}{\lambda' \mu} \Big)} e^{- \varrho_y \Big( \tfrac{\lambda'}{\mu (\lambda' + \mu)} \Big) } I_0 \Big( \frac{2}{\mu} \sqrt{\varrho_h \varrho_y} \Big) \ d\varrho_h \\[6pt] &= \frac{\lambda + \mu}{\lambda + \mu - t \lambda \mu} \cdot \exp \Big( \varrho_y \cdot \frac{t \lambda^2}{(\lambda + \mu) (\lambda + \mu (1-t \lambda))} \Big). \\[6pt] \end{aligned} \end{equation}

Taking the logarithm gives the cumulant generating function:

$$K(t) \equiv \log m(t) = \log(\lambda + \mu) - \log(\lambda + \mu - t \lambda \mu) + \varrho_y \cdot \frac{t \lambda^2}{(\lambda + \mu) (\lambda + \mu (1-t \lambda))}.$$

Mean: The first derivative of the cumulant generating function is:

$$\frac{dK}{dt}(t) = \frac{\lambda \mu}{\lambda + \mu - t \lambda \mu} + \varrho_y \cdot \frac{\lambda^2}{(\lambda + \mu) (\lambda + \mu (1-t \lambda))} + \varrho_y \cdot \frac{t \mu \lambda^3}{(\lambda + \mu) (\lambda + \mu (1-t \lambda))^2},$$

so we have:

$$\mathbb{E}( \varrho_h | \varrho_y ) = \frac{dK}{dt}(0) = \frac{\lambda \mu}{\lambda + \mu} + \varrho_y \cdot \frac{\lambda^2}{(\lambda + \mu)^2} = \frac{\lambda \mu}{\lambda + \mu} \Bigg( 1 + \varrho_y \cdot \frac{\lambda}{\mu (\lambda + \mu)} \Bigg).$$

Variance: The second derivative of the cumulant generating function is:

\begin{equation} \begin{aligned} \frac{d^2 K}{dt^2}(t) &= \Big( \frac{\lambda \mu}{\lambda + \mu - t \lambda \mu} \Big)^2 + \varrho_y \cdot \frac{2 \mu \lambda^3}{(\lambda + \mu) (\lambda + \mu (1-t \lambda))^2} \\[6pt] &\quad + \varrho_y \cdot \frac{t \mu^2 \lambda^4}{(\lambda + \mu) (\lambda + \mu (1-t \lambda))^3}, \end{aligned} \end{equation}

so we have:

$$\mathbb{V}( \varrho_h | \varrho_y ) = \frac{d^2 K}{dt^2}(0) = \Big( \frac{\lambda \mu}{\lambda + \mu} \Big)^2 + \varrho_y \cdot \frac{2 \mu \lambda^3}{(\lambda + \mu)^3} = \Big( \frac{\lambda \mu}{\lambda + \mu} \Big)^2 \Bigg( 1 + 2 \varrho_y \cdot \frac{\lambda}{\mu (\lambda + \mu)} \Bigg).$$

• in derivation of moment generating function m(t), how to solve the integral containing Bessel function? As the integrand is pdf, the integral has 1? – W W W Mar 2 '19 at 5:44
• That's correct - integral is the same pdf, but with parameter $\lambda'$ instead of $\lambda$, so it integrates to one. – Ben - Reinstate Monica Mar 2 '19 at 7:35