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From the [[ECE_600_Prerequisites|ECE600 Pre-requisites notes]] of [[user:han84|Sangchun Han]], [[ECE]] PhD student. | From the [[ECE_600_Prerequisites|ECE600 Pre-requisites notes]] of [[user:han84|Sangchun Han]], [[ECE]] PhD student. | ||
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− | The joint characteristic function of two joint-distributed RVs <math>\mathbf{X}</math> and <math>\mathbf{Y}</math> is | + | The joint characteristic function of two joint-distributed RVs <math class="inline">\mathbf{X}</math> and <math class="inline">\mathbf{Y}</math> is |
− | <math>\Phi_{\mathbf{XY}}\left(\omega_{1},\omega_{2}\right)\triangleq E\left[e^{i\left(\omega_{1}\mathbf{X}+\omega_{2}\mathbf{Y}\right)}\right]=\iint_{\mathbf{R}^{2}}e^{i\left(\omega_{1}\mathbf{X}+\omega_{2}\mathbf{Y}\right)}f_{\mathbf{XY}}\left(x,y\right)dxdy=\text{2-dim Fourier transform}.</math> | + | <math class="inline">\Phi_{\mathbf{XY}}\left(\omega_{1},\omega_{2}\right)\triangleq E\left[e^{i\left(\omega_{1}\mathbf{X}+\omega_{2}\mathbf{Y}\right)}\right]=\iint_{\mathbf{R}^{2}}e^{i\left(\omega_{1}\mathbf{X}+\omega_{2}\mathbf{Y}\right)}f_{\mathbf{XY}}\left(x,y\right)dxdy=\text{2-dim Fourier transform}.</math> |
Note | Note | ||
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Inverse Fourier transform relation: | Inverse Fourier transform relation: | ||
− | <math>f_{\mathbf{XY}}\left(x,y\right)=\frac{1}{\left(2\pi\right)^{2}}\iint_{\mathbf{R}^{2}}e^{-i\left(\omega_{1}x+\omega_{2}y\right)}\Phi_{\mathbf{XY}}\left(\omega_{1},\omega_{2}\right)d\omega_{1}d\omega_{2}.</math> | + | <math class="inline">f_{\mathbf{XY}}\left(x,y\right)=\frac{1}{\left(2\pi\right)^{2}}\iint_{\mathbf{R}^{2}}e^{-i\left(\omega_{1}x+\omega_{2}y\right)}\Phi_{\mathbf{XY}}\left(\omega_{1},\omega_{2}\right)d\omega_{1}d\omega_{2}.</math> |
Note | Note | ||
− | 1. <math>\Phi_{\mathbf{X}}\left(\omega\right)=\Phi_{\mathbf{XY}}\left(\omega,0\right)</math> and <math>\Phi_{\mathbf{Y}}\left(\omega\right)=\Phi_{\mathbf{XY}}\left(0,\omega\right)</math> . | + | 1. <math class="inline">\Phi_{\mathbf{X}}\left(\omega\right)=\Phi_{\mathbf{XY}}\left(\omega,0\right)</math> and <math class="inline">\Phi_{\mathbf{Y}}\left(\omega\right)=\Phi_{\mathbf{XY}}\left(0,\omega\right)</math> . |
− | 2. If <math>\mathbf{Z}=a\mathbf{X}+b\mathbf{Y}</math> , then <math>\Phi_{\mathbf{Z}}\left(\omega\right)=E\left[e^{i\omega\left(a\mathbf{X}+b\mathbf{Y}\right)}\right]=E\left[e^{i\left(\left(\omega a\right)\mathbf{X}+\left(\omega b\right)\mathbf{Y}\right)}\right]=\Phi_{\mathbf{XY}}\left(\omega a,\omega b\right)</math>. | + | 2. If <math class="inline">\mathbf{Z}=a\mathbf{X}+b\mathbf{Y}</math> , then <math class="inline">\Phi_{\mathbf{Z}}\left(\omega\right)=E\left[e^{i\omega\left(a\mathbf{X}+b\mathbf{Y}\right)}\right]=E\left[e^{i\left(\left(\omega a\right)\mathbf{X}+\left(\omega b\right)\mathbf{Y}\right)}\right]=\Phi_{\mathbf{XY}}\left(\omega a,\omega b\right)</math>. |
• This is used for sum of two Gaussian RVs. | • This is used for sum of two Gaussian RVs. | ||
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Theorem | Theorem | ||
− | If two jointly-distributed RVs <math>\mathbf{X}</math> and <math>\mathbf{Y}</math> are statistically independent, then <math>\Phi_{\mathbf{XY}}\left(\omega_{1},\omega_{2}\right)=\Phi_{\mathbf{X}}\left(\omega_{1}\right)\mathbf{\Phi}_{\mathbf{Y}}\left(\omega_{2}\right)</math> . | + | If two jointly-distributed RVs <math class="inline">\mathbf{X}</math> and <math class="inline">\mathbf{Y}</math> are statistically independent, then <math class="inline">\Phi_{\mathbf{XY}}\left(\omega_{1},\omega_{2}\right)=\Phi_{\mathbf{X}}\left(\omega_{1}\right)\mathbf{\Phi}_{\mathbf{Y}}\left(\omega_{2}\right)</math> . |
Proof | Proof | ||
− | <math>\Phi_{\mathbf{XY}}\left(\omega_{1},\omega_{2}\right)=E\left[e^{i\left(\omega_{1}\mathbf{X}+\omega_{2}\mathbf{Y}\right)}\right]=E\left[e^{i\omega_{1}\mathbf{X}}e^{i\omega_{2}\mathbf{Y}}\right]=E\left[e^{i\omega_{1}\mathbf{X}}\right]E\left[e^{i\omega_{2}\mathbf{Y}}\right]=\Phi_{\mathbf{X}}\left(\omega_{1}\right)\mathbf{\Phi}_{\mathbf{Y}}\left(\omega_{2}\right)</math>. | + | <math class="inline">\Phi_{\mathbf{XY}}\left(\omega_{1},\omega_{2}\right)=E\left[e^{i\left(\omega_{1}\mathbf{X}+\omega_{2}\mathbf{Y}\right)}\right]=E\left[e^{i\omega_{1}\mathbf{X}}e^{i\omega_{2}\mathbf{Y}}\right]=E\left[e^{i\omega_{1}\mathbf{X}}\right]E\left[e^{i\omega_{2}\mathbf{Y}}\right]=\Phi_{\mathbf{X}}\left(\omega_{1}\right)\mathbf{\Phi}_{\mathbf{Y}}\left(\omega_{2}\right)</math>. |
− | <math>\because \mathbf{X}</math> and <math>\mathbf{Y}</math> are statistically independent | + | <math class="inline">\because \mathbf{X}</math> and <math class="inline">\mathbf{Y}</math> are statistically independent |
− | <math>\Longrightarrow e^{i\omega_{1}\mathbf{X}}</math> and <math>e^{i\omega_{2}\mathbf{Y}}</math> are statistically independent | + | <math class="inline">\Longrightarrow e^{i\omega_{1}\mathbf{X}}</math> and <math class="inline">e^{i\omega_{2}\mathbf{Y}}</math> are statistically independent |
− | <math>\Longrightarrow e^{i\omega_{1}\mathbf{X}}</math> and <math>e^{i\omega_{2}\mathbf{Y}}</math> are uncorrelated. | + | <math class="inline">\Longrightarrow e^{i\omega_{1}\mathbf{X}}</math> and <math class="inline">e^{i\omega_{2}\mathbf{Y}}</math> are uncorrelated. |
Fact | Fact | ||
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The joint characteristic function of two jointly-distributed Gaussian RVs is | The joint characteristic function of two jointly-distributed Gaussian RVs is | ||
− | <math>\Phi_{\mathbf{XY}}\left(\omega_{1},\omega_{2}\right)=e^{i\left(\eta_{\mathbf{X}}\omega_{1}+\eta_{\mathbf{Y}}\omega_{2}\right)}\cdot e^{-\frac{1}{2}\left[\sigma_{\mathbf{X}}^{2}\omega_{1}^{2}+2r\sigma_{\mathbf{X}}\sigma_{\mathbf{Y}}\omega_{1}\omega_{2}+\sigma_{\mathbf{Y}}^{2}\omega_{2}^{2}\right]}</math>. | + | <math class="inline">\Phi_{\mathbf{XY}}\left(\omega_{1},\omega_{2}\right)=e^{i\left(\eta_{\mathbf{X}}\omega_{1}+\eta_{\mathbf{Y}}\omega_{2}\right)}\cdot e^{-\frac{1}{2}\left[\sigma_{\mathbf{X}}^{2}\omega_{1}^{2}+2r\sigma_{\mathbf{X}}\sigma_{\mathbf{Y}}\omega_{1}\omega_{2}+\sigma_{\mathbf{Y}}^{2}\omega_{2}^{2}\right]}</math>. |
− | If <math>\mathbf{Z}=\mathbf{X}+\mathbf{Y}</math> , then <math>\eta_{\mathbf{Z}}=\eta_{\mathbf{X}}+\eta_{\mathbf{Y}}</math> and <math>\sigma_{\mathbf{Z}}=\sqrt{\sigma_{\mathbf{X}}^{2}+2r\sigma_{\mathbf{X}}\sigma_{\mathbf{Y}}+\sigma_{\mathbf{Y}}^{2}}</math> because <math>\Phi_{\mathbf{Z}}\left(\omega\right)=\Phi_{\mathbf{XY}}\left(\omega,\omega\right)=e^{i\left(\eta_{\mathbf{X}}+\eta_{\mathbf{Y}}\right)\omega}\cdot e^{-\frac{1}{2}\left[\sigma_{\mathbf{X}}^{2}+2r\sigma_{\mathbf{X}}\sigma_{\mathbf{Y}}+\sigma_{\mathbf{Y}}^{2}\right]\omega^{2}}</math>. | + | If <math class="inline">\mathbf{Z}=\mathbf{X}+\mathbf{Y}</math> , then <math class="inline">\eta_{\mathbf{Z}}=\eta_{\mathbf{X}}+\eta_{\mathbf{Y}}</math> and <math class="inline">\sigma_{\mathbf{Z}}=\sqrt{\sigma_{\mathbf{X}}^{2}+2r\sigma_{\mathbf{X}}\sigma_{\mathbf{Y}}+\sigma_{\mathbf{Y}}^{2}}</math> because <math class="inline">\Phi_{\mathbf{Z}}\left(\omega\right)=\Phi_{\mathbf{XY}}\left(\omega,\omega\right)=e^{i\left(\eta_{\mathbf{X}}+\eta_{\mathbf{Y}}\right)\omega}\cdot e^{-\frac{1}{2}\left[\sigma_{\mathbf{X}}^{2}+2r\sigma_{\mathbf{X}}\sigma_{\mathbf{Y}}+\sigma_{\mathbf{Y}}^{2}\right]\omega^{2}}</math>. |
Joint Moment Generation Function | Joint Moment Generation Function | ||
− | The joint moment generating function of two jointly-distributed RVs <math>\mathbf{X}</math> and <math>\mathbf{Y}</math> is | + | The joint moment generating function of two jointly-distributed RVs <math class="inline">\mathbf{X}</math> and <math class="inline">\mathbf{Y}</math> is |
− | <math>\phi_{\mathbf{XY}}\left(s_{1},s_{2}\right)\triangleq E\left[e^{s_{1}\mathbf{X}+s_{2}\mathbf{Y}}\right]=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{s_{1}\mathbf{X}+s_{2}\mathbf{Y}}f_{\mathbf{XY}}\left(x,y\right)dxdy</math>. | + | <math class="inline">\phi_{\mathbf{XY}}\left(s_{1},s_{2}\right)\triangleq E\left[e^{s_{1}\mathbf{X}+s_{2}\mathbf{Y}}\right]=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{s_{1}\mathbf{X}+s_{2}\mathbf{Y}}f_{\mathbf{XY}}\left(x,y\right)dxdy</math>. |
Moment Theorem | Moment Theorem | ||
− | The joint noncentral moment <math>E\left[\mathbf{X}^{j}\mathbf{Y}^{k}\right]</math> is given by | + | The joint noncentral moment <math class="inline">E\left[\mathbf{X}^{j}\mathbf{Y}^{k}\right]</math> is given by |
− | <math>E\left[\mathbf{X}^{j}\mathbf{Y}^{k}\right]=\frac{\partial^{j}\partial^{k}}{\partial s_{1}^{j}\partial s_{2}^{k}}\left(\phi_{\mathbf{XY}}\left(s_{1},s_{2}\right)\right)|_{s_{1}=0,s_{2}=0}</math>. | + | <math class="inline">E\left[\mathbf{X}^{j}\mathbf{Y}^{k}\right]=\frac{\partial^{j}\partial^{k}}{\partial s_{1}^{j}\partial s_{2}^{k}}\left(\phi_{\mathbf{XY}}\left(s_{1},s_{2}\right)\right)|_{s_{1}=0,s_{2}=0}</math>. |
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[[ECE600|Back to ECE600]] | [[ECE600|Back to ECE600]] | ||
[[ECE 600 Prerequisites|Back to ECE 600 Prerequisites]] | [[ECE 600 Prerequisites|Back to ECE 600 Prerequisites]] |
Latest revision as of 10:32, 30 November 2010
1.11 Joint Characteristic Function
From the ECE600 Pre-requisites notes of Sangchun Han, ECE PhD student.
The joint characteristic function of two joint-distributed RVs $ \mathbf{X} $ and $ \mathbf{Y} $ is
$ \Phi_{\mathbf{XY}}\left(\omega_{1},\omega_{2}\right)\triangleq E\left[e^{i\left(\omega_{1}\mathbf{X}+\omega_{2}\mathbf{Y}\right)}\right]=\iint_{\mathbf{R}^{2}}e^{i\left(\omega_{1}\mathbf{X}+\omega_{2}\mathbf{Y}\right)}f_{\mathbf{XY}}\left(x,y\right)dxdy=\text{2-dim Fourier transform}. $
Note
Inverse Fourier transform relation:
$ f_{\mathbf{XY}}\left(x,y\right)=\frac{1}{\left(2\pi\right)^{2}}\iint_{\mathbf{R}^{2}}e^{-i\left(\omega_{1}x+\omega_{2}y\right)}\Phi_{\mathbf{XY}}\left(\omega_{1},\omega_{2}\right)d\omega_{1}d\omega_{2}. $
Note
1. $ \Phi_{\mathbf{X}}\left(\omega\right)=\Phi_{\mathbf{XY}}\left(\omega,0\right) $ and $ \Phi_{\mathbf{Y}}\left(\omega\right)=\Phi_{\mathbf{XY}}\left(0,\omega\right) $ .
2. If $ \mathbf{Z}=a\mathbf{X}+b\mathbf{Y} $ , then $ \Phi_{\mathbf{Z}}\left(\omega\right)=E\left[e^{i\omega\left(a\mathbf{X}+b\mathbf{Y}\right)}\right]=E\left[e^{i\left(\left(\omega a\right)\mathbf{X}+\left(\omega b\right)\mathbf{Y}\right)}\right]=\Phi_{\mathbf{XY}}\left(\omega a,\omega b\right) $.
• This is used for sum of two Gaussian RVs.
Theorem
If two jointly-distributed RVs $ \mathbf{X} $ and $ \mathbf{Y} $ are statistically independent, then $ \Phi_{\mathbf{XY}}\left(\omega_{1},\omega_{2}\right)=\Phi_{\mathbf{X}}\left(\omega_{1}\right)\mathbf{\Phi}_{\mathbf{Y}}\left(\omega_{2}\right) $ .
Proof
$ \Phi_{\mathbf{XY}}\left(\omega_{1},\omega_{2}\right)=E\left[e^{i\left(\omega_{1}\mathbf{X}+\omega_{2}\mathbf{Y}\right)}\right]=E\left[e^{i\omega_{1}\mathbf{X}}e^{i\omega_{2}\mathbf{Y}}\right]=E\left[e^{i\omega_{1}\mathbf{X}}\right]E\left[e^{i\omega_{2}\mathbf{Y}}\right]=\Phi_{\mathbf{X}}\left(\omega_{1}\right)\mathbf{\Phi}_{\mathbf{Y}}\left(\omega_{2}\right) $.
$ \because \mathbf{X} $ and $ \mathbf{Y} $ are statistically independent
$ \Longrightarrow e^{i\omega_{1}\mathbf{X}} $ and $ e^{i\omega_{2}\mathbf{Y}} $ are statistically independent
$ \Longrightarrow e^{i\omega_{1}\mathbf{X}} $ and $ e^{i\omega_{2}\mathbf{Y}} $ are uncorrelated.
Fact
The joint characteristic function of two jointly-distributed Gaussian RVs is
$ \Phi_{\mathbf{XY}}\left(\omega_{1},\omega_{2}\right)=e^{i\left(\eta_{\mathbf{X}}\omega_{1}+\eta_{\mathbf{Y}}\omega_{2}\right)}\cdot e^{-\frac{1}{2}\left[\sigma_{\mathbf{X}}^{2}\omega_{1}^{2}+2r\sigma_{\mathbf{X}}\sigma_{\mathbf{Y}}\omega_{1}\omega_{2}+\sigma_{\mathbf{Y}}^{2}\omega_{2}^{2}\right]} $.
If $ \mathbf{Z}=\mathbf{X}+\mathbf{Y} $ , then $ \eta_{\mathbf{Z}}=\eta_{\mathbf{X}}+\eta_{\mathbf{Y}} $ and $ \sigma_{\mathbf{Z}}=\sqrt{\sigma_{\mathbf{X}}^{2}+2r\sigma_{\mathbf{X}}\sigma_{\mathbf{Y}}+\sigma_{\mathbf{Y}}^{2}} $ because $ \Phi_{\mathbf{Z}}\left(\omega\right)=\Phi_{\mathbf{XY}}\left(\omega,\omega\right)=e^{i\left(\eta_{\mathbf{X}}+\eta_{\mathbf{Y}}\right)\omega}\cdot e^{-\frac{1}{2}\left[\sigma_{\mathbf{X}}^{2}+2r\sigma_{\mathbf{X}}\sigma_{\mathbf{Y}}+\sigma_{\mathbf{Y}}^{2}\right]\omega^{2}} $.
Joint Moment Generation Function
The joint moment generating function of two jointly-distributed RVs $ \mathbf{X} $ and $ \mathbf{Y} $ is
$ \phi_{\mathbf{XY}}\left(s_{1},s_{2}\right)\triangleq E\left[e^{s_{1}\mathbf{X}+s_{2}\mathbf{Y}}\right]=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{s_{1}\mathbf{X}+s_{2}\mathbf{Y}}f_{\mathbf{XY}}\left(x,y\right)dxdy $.
Moment Theorem
The joint noncentral moment $ E\left[\mathbf{X}^{j}\mathbf{Y}^{k}\right] $ is given by
$ E\left[\mathbf{X}^{j}\mathbf{Y}^{k}\right]=\frac{\partial^{j}\partial^{k}}{\partial s_{1}^{j}\partial s_{2}^{k}}\left(\phi_{\mathbf{XY}}\left(s_{1},s_{2}\right)\right)|_{s_{1}=0,s_{2}=0} $.