应用随机过程.md

第三章 平稳随机过程习题

1.设随机过程$X(t) = Acos(\omega_0 t + \Phi), -\infin < t < +\infin$, 其中$\omega_0$是常数,$A$与$\Phi$是独立随机变量.$\Phi$服从在区间$\left(0,2\pi\right)$中的均匀分布.$A$服从瑞利分布，其密度为 $$ f(x) = \left{\begin{array}{l}\frac{x}{\sigma^2}e^{-\frac{x^2}{2\sigma^2}}, x\ge 0\0,x<0\end{array}\right. $$ 又设随机过程$Y(t)=Bcos\omega_0t+Csin\omega_0t,-\infin<t<+\infin$,其中$B$和$C$是相互独立的正态变量，且都具有分布$N(0,\sigma^2)$.

(1). 试证$X(t)$是平稳过程 $$ \begin{eqnarray} m_\mathbf{X} &=& EX(t) =EAcos(\omega_0t+\Phi) \overset{A和\Phi相互独立}{=} EAEcos(\omega_0t+\Phi) \nonumber\ &=& EA\int^{2\pi}{0}cos(\omega_0t+\varphi)\frac{1}{2\pi}d\varphi= 0 \nonumber\ R\mathbf{X}(t,t +\tau)&=& EX(t)X(t+\tau)=EA^2Ecos(\omega_0t+\Phi)cos(\omega_0t+\omega_0\tau+\Phi) \nonumber \ &=& \int^{+\infin}{0}x^2\frac{x}{\sigma^2}e^{-\frac{x^2}{2\sigma^2}}dx\int^{2\pi}{0}cos(\omega_0t+\varphi)cos(\omega_0t+\omega_0\tau+\varphi)\frac{1}{2\pi}d\varphi\nonumber\ &\overset{分部积分+三角变换}{=}&\sigma^2cos\omega_0\tau \end{eqnarray} $$ $m_\mathbf{X}$为常数且$R_\mathbf{X}(t,t +\tau)$与$t$无关，且仅和$\tau$ 有关，故而$X(t)$为平稳随机过程。

(2).用本章$\S1$中的例题说明$Y(t)$是平稳过程 $$ \begin{eqnarray} m_{X} &=& EY(t) = EBcos\omega_0t+ECsin\omega_0t = 0\ R_Y(t,t+\tau) &=& EY(t)Y(t+\tau)\nonumber\ &=&E[(Bcos\omega_0t+Csin\omega_0t)(Bcos\omega_0(t+\tau)+Csin\omega_0(t+\tau)] \nonumber \ &\overset{B,C相互独立}{=}&EB^2cos\omega_0tcos\omega_0(t+\tau)+EC^2sin\omega_0tsin\omega_0(t+\tau)\nonumber\ &=& \sigma^2cos\omega_0\tau \end{eqnarray} $$ $m_\mathbf{Y}$为常数且$R_\mathbf{Y}(t,t +\tau)$与$t$无关，且仅和$\tau$ 有关，故而$Y(t)$为平稳随机过程。

(3).如果$Y(t)=Bcos\omega_0t+Csin\omega_0t,-\infin<t<+\infin$,其中$B=Acos\Phi,C=-Asin\Phi$.试证$B$与$C$是分别具有$N(0,\sigma^2)$的独立正态变量. $$ \left{\begin{array}{l}B=Acos\Phi\C=-Asin\Phi\end{array}\right. $$ 已知$A$和$\Phi$相互独立，且$A$和$\Phi$的分布已知，可推出: $$ \left{\begin{array}{l}A=\sqrt{B^2+C^2}\\Phi=-\tan^{-1}\frac{B}{C}\end{array}\right. $$ $A$和$\Phi$的联合分布为: $$ p(a,\varphi) = p_A(a)\space\bull \space p_\Phi (\varphi) = \frac{a}{\sigma^2}e^{-\frac{a^2}{2\sigma^2}}\frac{1}{2\pi} $$ 令$B$和$C$的联合分布密度为$q(b,c)$，则有: $$ q(b,c) = q(a,\varphi)\space\bull\space \left|\mathbf{J}\right| $$ 其中 $$ \left|\mathbf{J}\right| = \begin{vmatrix} \frac{\partial a}{\partial b} & \frac{\partial a}{\partial c} \ \frac{\partial \varphi}{\partial b} & \frac{\partial \varphi}{\partial c} \ \end{vmatrix} = \left|-\frac{1}{\sqrt{b^2+c^2}}\right| = \left|-\frac{1}{a}\right|\ $$ 故而: $$ q(b,c) = \frac{a}{\sigma^2}e^{-\frac{a^2}{2\sigma^2}}\frac{1}{2\pi} \frac{1}{a}=\frac{1}{\sigma^2}e^{-\frac{a^2}{2\sigma^2}}\frac{1}{2\pi} = \frac{1}{\sigma^2}e^{-\frac{b^2+c^2}{2\sigma^2}}\frac{1}{\left(\sqrt{2\pi}\right)^2}\ =\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{b^2}{2\sigma^2}}\bull\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{c^2}{2\sigma^2}} = q_B(b)q_C(c) $$ 所以$B$与$C$是分别具有$N(0,\sigma^2)$的独立正态变量。

2.设随机过程$X(t) = Acos(\omega_0 t + \Phi), -\infin < t < +\infin$, 其中$\omega_0$是常数,$A$与$\Phi$是独立随机变量.$\Phi$服从在区间$\left(0,2\pi\right)$中的均匀分布.试问$X(t)$是否具有各态历经性。 $$ m_X = EX(t) = 0 \ R_X(\tau) = EA^2\bull \frac{1}{2}cos\omega_0\tau $$ 故而$X(t)$是宽平稳随机过程.从而 $$ \begin{eqnarray} <X(t)> &=& \mathop {{\mathop{\rm l}\nolimits} \cdot i \cdot m}\limits_{T \to \infty }\frac{1}{2T}\int^{T}{-T}X(t)dt = \mathop {{\mathop{\rm l}\nolimits} \cdot i \cdot m}\limits{T \to \infty }\frac{1}{2T}\int^{T}{-T}Acos(\omega_0 t + \Phi)dt \nonumber \ &=&\mathop {{\mathop{\rm l}\nolimits} \cdot i \cdot m}\limits{T \to \infty }\frac{A}{2T}\int^{T}{-T}(cos(\omega_0 t)cos\Phi - sin(\omega_0 t)sin\Phi)dt \nonumber \ &=& \mathop {{\mathop{\rm l}\nolimits} \cdot i \cdot m}\limits{T \to \infty }\frac{Acos\Phi}{T\omega_0}sin\omega_0t = \mathop {{\mathop{\rm l}\nolimits} \cdot i \cdot m}\limits_{T \to \infty }\frac{sin\omega_0t}{T\omega_0}Acos\Phi = 0 \end{eqnarray} $$ 所以$E(t)=<X(t)>$,即$X(t)$满足期望的各态历性,从而有 $$ \begin{eqnarray} <X(t)X(t+\tau)> &=& \mathop {{\mathop{\rm l}\nolimits} \cdot i \cdot m}\limits_{T \to \infty }\frac{A^2}{2T}\int^{T}{-T}cos(\omega_0t+\Phi)cos(\omega_0t+\omega\tau+\Phi)dt \nonumber \ &=& \mathop {{\mathop{\rm l}\nolimits} \cdot i \cdot m}\limits{T \to \infty }\frac{A^2}{2T}\int^{T}{-T}[cos(2\omega_0t+2\Phi + \omega\tau)+cos(\omega\tau)]dt \nonumber \ &=&\mathop {{\mathop{\rm l}\nolimits} \cdot i \cdot m}\limits{T \to \infty }\frac{A^2}{2T}Tcos(\omega\tau) = \frac{A^2}{2}cos\omega_0\tau \ne R_X(\tau)\nonumber \ \end{eqnarray} $$ 所以$X(t)$的相关函数不满足各态历性。

3.设随机过程$X(t) = Asint + Bcost, -\infin < t < +\infin$,其中$A,B$是均值为0的不相关随机变量，$EA^2=EB^2$.证$X(t)$的各态历经性。 $$ m_X=EX(t) = EAsint + EBcost = sintEA +costEB = 0 $$

$$ \begin{eqnarray} R_X(t,t+\tau) &=& E[(Asint+Bcost)(Asint(t+\tau)+Bcos(t+\tau))] \nonumber \\ &=& EA^2sintsin(t+\tau) + EB^2costcos(t+\tau) \nonumber \\ &=& (cos(t+\tau -t))EA^2 = cos\tau EA^2 \end{eqnarray} $$

$m_X$为常数，$R_X(t,t+\tau)$与$t$无关，仅与$\tau$相关，所以$X(t)$是宽平稳随机过程。 $$ \begin{eqnarray} <X(t)> &=& \mathop {{\mathop{\rm l}\nolimits} \cdot i \cdot m}\limits_{T \to \infty }\frac{1}{2T}\int^T_{-T}(Asint+Bcost)dt = \mathop {{\mathop{\rm l}\nolimits} \cdot i \cdot m}\limits_{T \to \infty }\frac{B}{2T}\int^T_{-T}costdt \nonumber \ &=& 0 \end{eqnarray} $$

$$ \begin{eqnarray} <X(t)X(t+\tau)> &=& \mathop {{\mathop{\rm l}\nolimits} \cdot i \cdot m}\limits_{T \to \infty }\frac{1}{2T}\int^T_{-T}(Asint+Bcost)(Asin(t+\tau)+Bcos(t+\tau))dt \nonumber \\ &=& \mathop {{\mathop{\rm l}\nolimits} \cdot i \cdot m}\limits_{T \to \infty }\frac{1}{2T}\int^T_{-T}[A^2sintsin(t+\tau)+ABsin(2t+\tau)+B^2costcos(t+\tau)]dt \nonumber \\ &=& \mathop {{\mathop{\rm l}\nolimits} \cdot i \cdot m}\limits_{T \to \infty }\frac{1}{2T}\int^T_{-T}[A^2\frac{cos\tau - cos(2t+\tau)}{2} + ABsin(2t+\tau) +B^2\frac{cos\tau+cos(2t+\tau)}{2}]dt \nonumber \\ &=& \frac{A^2+B^2}{2}cos\tau \ne R_X(t,t+\tau) \end{eqnarray} $$

故而$X(t)$期望具有各态历经性，相关函数无各态历经性。

4.设平稳过程${X(t),-\infin < t < +\infin}$的相关函数为$R_X(\tau)=Ae^{-a|\tau|}(1+a|\tau|)$,其中$A,a$都是正常数,而$EX(t)=0$.试问$X(t)$对数学期望是否有各态历性?

由定理1(数学期望各态历经定理)知 $$ <X(t)> = \mathop {{\mathop{\rm l}\nolimits} \cdot i \cdot m}\limits_{T \to \infty }\frac{1}{2T}\int^T_{-T}X(t)dt = m_X, a.s \Leftrightarrow \nonumber \ \mathop{{\mathop{\rm l}\nolimits}im}\limits_{T \to \infty }\frac{1}{T}\int_0^{2T}(1-\frac{\tau}{2T})(R_X(\tau)-m^2_X)d\tau =0 $$ 故而只需证明下式成立 $$ \mathop{{\mathop{\rm l}\nolimits}im}\limits_{T \to \infty }\frac{1}{T}\int_0^{2T}(1-\frac{\tau}{2T})(R_X(\tau)-m^2_X)d\tau =0 $$ 推演过程如下: $$ \begin{eqnarray} 原式 &=& \mathop{{\mathop{\rm l}\nolimits}im}\limits_{T \to \infty }\frac{1}{T}\int_0^{2T}(1-\frac{\tau}{2T})Ae^{-a|\tau|}(1+a|\tau|)d\tau \nonumber \ &=& \mathop{{\mathop{\rm l}\nolimits}im}\limits_{T \to \infty }\frac{A}{aT}\int_0^{2T}(1-\frac{x}{2aT})e^{-x}(1+x)dx \nonumber \ &\overset{x^ne^{-x}形式的积分}{=}& 0 \end{eqnarray} $$ 所以$X(t)$对数学期望有各态历性。

5.设随机过程$X(t) = Acos(\omega t + \Phi), -\infin < t < +\infin$，其中$\omega,A$与$\Phi$是独立随机变量.$A$均值为2,方差为4.$\Phi$在$(-\pi,\pi)$上服从均匀分布，$\omega$在$(-5,5)$上服从均匀分布，求$X(t)$的自相关函数，并讨论平稳性及各态历经性。 $$ \begin{eqnarray} m_X &=& EX(t) = EAEcos(\omega t+\Phi) \nonumber \ &=& EA(Ecos\omega tEcos\Phi - Esin\omega tEsin\Phi) \nonumber \ &=& 0 \end{eqnarray} $$

$$ R_X(t,t+\tau) = EA^2Ecost(\omega t+\Phi)cos(\omega t+\omega\tau + \Phi) = EA^2Ecos\omega\tau=\frac{4sin5\tau}{5\tau} $$

所以$X(t)$是平稳随机过程。

$$ <X(t)> = \mathop {{\mathop{\rm l}\nolimits} \cdot i \cdot m}\limits_{T \to \infty }\frac{1}{2T}\int^{T}{-T}Acos(\omega t+\Phi)dt = \mathop {{\mathop{\rm l}\nolimits} \cdot i \cdot m}\limits{T \to \infty }\frac{Acos\Phi}{\omega}\frac{sin\omega T}{T} = 0 = m_X $$

所以$X(t)$对数学期望有各态历经性。 $$ \begin{eqnarray} <X(t)X(t+\tau)>&=&\mathop {{\mathop{\rm l}\nolimits} \cdot i \cdot m}\limits_{T \to \infty }\frac{1}{2T}\int^{T}{-T}A^2cos(\omega t+\Phi)cos(\omega t+\omega\tau+\Phi)dt \nonumber \ &=& \mathop {{\mathop{\rm l}\nolimits} \cdot i \cdot m}\limits{T \to \infty }\frac{A^2}{4T}\int^{T}_{-T}[cos(2\omega t+2\Phi + \omega\tau)+cos(\omega\tau)]dt \nonumber \ &=&\frac{A^2cost\omega\tau}{2T} \ne R_X(t,t+\tau) \end{eqnarray} $$ 所以$X(t)$对相关函数不具有各态历经性。

6.设$X(t)$,$N(t)$平稳相关，$a\ll 1$. $Y(t) = aX(t-\tau_1) + N(t)$,求互相关函数$R_{XY}(\tau)$. 若$EN(t) = 0$且与$X(t)$相互独立，求$R_{XY}(\tau)$.
$$ \begin{eqnarray} R_{XY}(\tau) &=& E[X(t)Y(t+\tau)] = E[aX(t)X(t+\tau-\tau_1)+X(t)N(t+\tau)] \nonumber \ &=& aR_X(\tau-\tau_1)+R_{XN}(\tau) \end{eqnarray} $$ 若若$EN(t) = 0$且与$X(t)$相互独立： $$ |R_{XN}(\tau)| \le \sqrt{R_x(0)}\sqrt{R_N(0)} = 0 \ \Rightarrow R_{XN}(\tau) =0 \Rightarrow R_{XY}(\tau) = aR_X(\tau-\tau_1) $$

7.两平稳过程$X(t) = acos(\omega_0 t + \Phi), Y(t)=bsin(\omega_0t+\Phi), -\infin < t < +\infin$其中$a,b,\omega_0$为常量，$\Phi \sim U(0,2\pi)$,求$R_{XY}(\tau), R_{YX}(\tau)$. $$ \begin{eqnarray} R_{XY}(\tau) &=& E[X(t)Y(t+\tau)] = abEcos(\omega_0t+\Phi)sin(\omega_0t+\omega_0\tau+\Phi) \nonumber \ &=& \frac{ab}{2}E[sin(2\omega_0t+2\Phi+\omega_0\tau) + sin(\omega_0\tau)] \nonumber \ &\overset{简单积分}{=}& \frac{ab}{2}sin\omega_0\tau \end{eqnarray} $$

$$ \begin{eqnarray} R_{YX}(\tau) &=& E[Y(t)X(t+\tau)] = abEsin(\omega_0t+\Phi)cos(\omega_0t+\omega_0\tau+\Phi) \nonumber \\ &=& \frac{ab}{2}E[sin(2\omega_0t+2\Phi+\omega_0\tau) + sin(-\omega_0\tau)] \nonumber \\ &\overset{简单积分}{=}& -\frac{ab}{2}sin\omega_0\tau \end{eqnarray}\\ R_{XY}(-\tau) = R_{YX}(\tau) $$

8.设${X(t),-\infin<t<+\infin}$是平稳过程.

(1)若存在$T>0$,使得$R_X(T)=R_X(0)$,则对固定$t$,有$X(t+T)=X(t), a.s.$ $$ E|X(t+T)-X(t)|^2=R_X(t+T,t+T)-2R_X(t,t+T) + R_X(t,t) = 2(R_X(0)-R_X(T)) = 0 \ \Rightarrow X(t+T) = X(t), a.s $$ (2)若$X(t)$可导,$E[X(t)X^\prime(t)]=R^\prime_X(0)=0$ $$ \begin{eqnarray} E[X(t)X^\prime(t)] &=& E[X(t)\mathop {{\mathop{\rm l}\nolimits} \cdot i \cdot m}\limits_{h \to 0 }\frac{X(t+h)-X(t)}{h}] \nonumber \ &=& \mathop {{\mathop{\rm l}\nolimits}i m}\limits_{h \to 0 }\frac{EX(t)X(t+h)-EX(t)X(t)}{h} \nonumber \ &=& \mathop {{\mathop{\rm l}\nolimits} i m}\limits_{h \to 0 }\frac{R_X(h)-R_X(0)}{h} = R^\prime_X(0) \overset{R_X(\tau)与t无关}{=} 0 \end{eqnarray} $$ (3)若$X(t)$可导,则$X^\prime(t)$是平稳过程,且其相关函数$R_{X^\prime}(\tau)=-\frac{\mathrm{d}^2R_X(\tau)}{\mathrm{d}\tau^2}$. $$ \begin{eqnarray} m_{X^\prime} &=& EX^\prime(t) =E\mathop {{\mathop{\rm l}\nolimits} \cdot i \cdot m}\limits_{h \to 0 }\frac{X(t+h)-X(t)}{h} \nonumber \ &=& \mathop {{\mathop{\rm l}\nolimits} i m}\limits_{h \to 0 }\frac{EX(t+h)-EX(t)}{h} \nonumber \ &=& m^\prime_X \space (与t无关的常数) \end{eqnarray} $$

$$ \begin{eqnarray} E[X^\prime(t)X^\prime(t+\tau)] &=& E\left[\mathop {{\mathop{\rm l}\nolimits} \cdot i \cdot m}\limits_{h_1 \to 0 }\frac{X(t+h_1)-X(t)}{h_1}\mathop {{\mathop{\rm l}\nolimits} \cdot i \cdot m}\limits_{h_2 \to 0 }\frac{X(t+\tau+h_2)-X(t+\tau)}{h_2}\right] \nonumber \\ &=& \mathop {{\mathop{\rm l}\nolimits}i m}\limits_{h_1 \to 0 }\mathop {{\mathop{\rm l}\nolimits}i m}\limits_{h_2 \to 0 }\frac{R_X(t+h_1,t+\tau+h_2)-R_X(t,t+\tau+h_2)-R_X(t+h_1,t+\tau)+R_X(t,t+\tau)}{h_1h_2} \nonumber \\ &=& \frac{\part^2}{\part t_2\part t_1}R_X(t_1,t_2) \overset{\tau=t2-t1}{=}R_{X^\prime}(\tau)=-\frac{\mathrm{d}^2R_X(\tau)}{\mathrm{d}\tau^2} \end{eqnarray} $$

**9.**设${X(t),-\infin<t<+\infin}$,${Y(t),-\infin<t<+\infin$是平稳过程.满足$Y^\prime(t)+aY(t)=X(t)$,$a$为非零常数，求证$m_Y=\frac{1}{a}m_X$. $$ E[Y^\prime(t)+aY(t)] = E[Y^\prime]+aE[Y(t)] = m_{Y^\prime}+am_Y = am_Y = EX(t) = m_X\ \Rightarrow m_Y=\frac{1}{a}m_X $$

**10.**设${X(t),-\infin<t<+\infin}$是平稳过程.$EX(t)=1,R(\tau)=1+e^{-2|\tau|}$.求随机变量$S=\int^1_0X(t)dt$的数学期望和方差。

由均方积分性质2:$E[\int^b_af(t)X(t)dt]=\int_a^bf(t)EX(t)dt=\int_a^bf(t)m_Xdt$知: $$ E[S] = E[\int^1_01\bull X(t)dt] = \int^1_01\bull EX(t)dt = t|^1_0 = 1 $$ 由均方积分性质3:$E|\int^b_af(t)X(t)dt|^2=\int_a^b\int_a^bf(s)f(t)R_X(s,t)dsdt$知: $$ \begin{eqnarray} DS &=& ES^2 -(ES)^2 = E\left|\int_0^11\bull X(t)dt\right|^2 -1 \nonumber \ &=& \int_0^1\int_0^11\bull1\bull R_X(s,t)dsdt - 1 \nonumber \ &=& \int_0^1\int_0^1e^{-2|t-s|}dsdt = \frac{1+e^{-2}}{2} \end{eqnarray} $$

**11.**设复随机过程$Z(t)=e^{i(\omega_0t+\Phi),-\infin<t<+\infin}$,$\Phi$是在$(0,2\pi)$上均匀分布的随机变量，$\omega_0$为常量.求$Z(t)$的相关函数，并讨论它的平稳性.

NOTE:题目貌似有问题,如果是$i(\omega_0t+\Phi)$的话$EZ(t)$就不是常数了。相关函数为$e^{i\tau}$

**13.**下列函数哪些是功率谱密度，哪些不是？

功率谱密度函数是非负，实的偶函数。

**14.**已知自相关函数求功率谱密度

(1)$R_X(\tau)=e^{-a|\tau|}cos\omega_0\tau$ $$ \begin{eqnarray} S_X(\omega) &=& \int^{+\infin}{-\infin}R_X(\tau)e^{-i\omega\tau}d\tau \nonumber \ &=& \int^{+\infin}{-\infin} \frac{1}{2}e^{-a|\tau|}[e^{i(\omega_0-\omega)\tau}+e^{-i(\omega_0+\omega)\tau}]d\tau \nonumber \ &=& \int^{+\infin}{-\infin} \frac{1}{2}e^{-a|\tau|}e^{i(\omega_0-\omega)\tau}d\tau + \int^{+\infin}{-\infin} \frac{1}{2}e^{-a|\tau|}e^{-i(\omega_0+\omega)\tau}d\tau \nonumber \ &\overset{傅立叶变换表}{=}& \frac{a}{a^2+(\omega_0-\omega)^2} + \frac{a}{a^2+(\omega_0+\omega)^2} \end{eqnarray} $$ (2)$R_X(\tau) = \left{\begin{array}{l}1-\frac{|\tau|}{T}, |\tau| \le T\0, other\end{array}\right.$ $$ S_X(\omega) \overset{傅立叶变换表}{=} \frac{4sin^2\frac{\omega T}{2}}{T\omega^2} $$ (3)$R_X(\tau)=4e^{-|\tau|}cos\pi\tau+cos3\pi\tau$ $$ S_X(\omega) = \mathcal{F}{4e^{-|\tau|}cos\pi\tau+cos3\pi\tau} = \mathcal{F}{4e^{-|\tau|}cos\pi\tau}+\mathcal{F}{cos3\pi\tau}\ =4[\frac{1}{1+(\omega+\pi)^2}+\frac{1}{1+(\omega-\pi)^2}]+\pi[\delta(\omega-3\pi)+\delta(\omega+3\pi)] $$ (4).$R_X(\tau)=\sigma^2e^{-a|\tau|}(cosb\tau-ab^{-1}sinb|\tau|)$ $$ \begin{eqnarray} S_X(\omega) &=& \mathcal{F}{\sigma^2e^{-a|\tau|}cosb\tau} - \mathcal{F}{ab^{-1}\sigma^2e^{-a|\tau|}sinb|\tau|} \ &=& \sigma^2[\frac{a}{a^2+(\omega+b)^2}+\frac{a}{a^2+(\omega-b)^2}] -ab^{-1}\sigma^2\mathcal{F}{\frac{1}{2j}e^{-a|\tau|}[e^{jb|\tau|}-e^{-jb|\tau|}]} \ &=& \sigma^2[\frac{a}{a^2+(\omega+b)^2}+\frac{a}{a^2+(\omega-b)^2}] -\frac{1}{2j}ab^{-1}\sigma^2\mathcal{F}{e^{(jb-a)|\tau|}-e^{(-jb-a)|\tau|}} \ &=& \sigma^2[\frac{a}{a^2+(\omega+b)^2}+\frac{a}{a^2+(\omega-b)^2}] - \frac{2a\sigma^2(a^2+b^2-\omega^2)}{a^2-b^2+\omega^2+4a^2b^2} \end{eqnarray} $$