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=Example. Two jointly distributed random variables (Joint characteristic function)=
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=Example. Two jointly distributed independent random variables=
  
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Let <math>\mathbf{X}</math>  and <math>\mathbf{Y}</math>  be tweo jointly distributed random variables having joint characteristic function
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Let <math class="inline">\mathbf{X}</math>  and <math class="inline">\mathbf{Y}</math>  be two jointly distributed, independent random variables. The pdf of <math class="inline">\mathbf{X}</math>  is
  
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<math>\Phi_{\mathbf{XY}}\left(\omega_{1},\omega_{2}\right)=\frac{1}{\left(1-i\omega_{1}\right)\left(1-i\omega_{2}\right)}.</math>  
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<math class="inline">f_{\mathbf{X}}\left(x\right)=xe^{-x^{2}/2}\cdot\mathbf{1}_{\left[0,\infty\right)}\left(x\right)</math>, and <math class="inline">\mathbf{Y}</math>  is a Gaussian random variable with mean 0  and variance 1 . Let <math class="inline">\mathbf{U}</math>  and <math class="inline">\mathbf{V}</math>  be two new random variables defined as <math class="inline">\mathbf{U}=\sqrt{\mathbf{X}^{2}+\mathbf{Y}^{2}}</math> and <math class="inline">\mathbf{V}=\lambda\mathbf{Y}/\mathbf{X}</math>  where <math class="inline">\lambda</math>  is a positive real number.
  
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(a) Calculate <math>E\left[\mathbf{X}\right]</math> .
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(a)
  
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<math>\Phi_{\mathbf{X}}\left(\omega\right)=\Phi_{\mathbf{XY}}\left(\omega,0\right)=\frac{1}{1-i\omega}=\left(1-i\omega\right)^{-1}</math>  
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Find the joint pdf of <math class="inline">\mathbf{U}</math>  and <math class="inline">\mathbf{V}</math> . (Direct pdf method)
  
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<math>E\left[\mathbf{X}\right]=\frac{d}{d\left(i\omega\right)}\Phi_{\mathbf{X}}\left(\omega\right)|_{i\omega=0}=(-1)(1-i\omega)^{-2}(-1)|_{i\omega=0}=1</math>  
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<math class="inline">f_{\mathbf{UV}}\left(u,v\right)=f_{\mathbf{XY}}\left(x\left(u,v\right),y\left(u,v\right)\right)\left|\frac{\partial\left(x,y\right)}{\partial\left(u,v\right)}\right|</math>  
  
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(b) Calculate <math>E\left[\mathbf{Y}\right]</math>  
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Solving for x  and y  in terms of u  and v , we have <math class="inline">u^{2}=x^{2}+y^{2}</math>  and <math class="inline">v^{2}=\frac{\lambda^{2}y^{2}}{x^{2}}\Longrightarrow y^{2}=\frac{v^{2}x^{2}}{\lambda^{2}}</math> .
  
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<math>E\left[\mathbf{Y}\right]=1</math>  
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Now, <math class="inline">u^{2}=x^{2}+y^{2}=x^{2}+\frac{v^{2}x^{2}}{\lambda^{2}}=x^{2}\left(1+v^{2}/\lambda^{2}\right)\Longrightarrow x=\frac{u}{\sqrt{1+v^{2}/\lambda^{2}}}\Longrightarrow x\left(u,v\right)=\frac{u}{\sqrt{1+v^{2}/\lambda^{2}}}</math> .
  
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(c) Calculate <math>E\left[\mathbf{XY}\right]</math> .
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Thus, <math class="inline">y=\frac{vx}{\lambda}=\frac{vu}{\lambda\sqrt{1+v^{2}/\lambda^{2}}}\Longrightarrow y\left(u,v\right)=\frac{vu}{\lambda\sqrt{1+v^{2}/\lambda^{2}}}</math> .
  
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<math>E\left[\mathbf{XY}\right]=\frac{\partial^{2}}{\partial\left(i\omega_{1}\right)\partial\left(i\omega_{2}\right)}\Phi_{\mathbf{XY}}\left(\omega_{1},\omega_{2}\right)|_{i\omega_{1}=i\omega_{2}=0}=\left(1-i\omega_{1}\right)^{-2}\left(1-i\omega_{2}\right)^{-2}|_{i\omega_{1}=i\omega_{2}=0}=1</math>
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Computing the Jacobian.
  
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(d) Calculate <math>E\left[\mathbf{X}^{j}\mathbf{Y}^{k}\right]</math> .
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<math class="inline">\frac{\partial\left(x,y\right)}{\partial\left(u,v\right)} =\left|\begin{array}{ll}
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\frac{\partial x}{\partial u}  \frac{\partial x}{\partial v}\\
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\frac{\partial y}{\partial u}  \frac{\partial y}{\partial v}
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\end{array}\right|=\left|\begin{array}{cc}
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\frac{1}{\sqrt{1+v^{2}/\lambda^{2}}}  \frac{-uv}{\lambda^{2}\left(1-v^{2}/\lambda^{2}\right)^{\frac{3}{2}}}\\
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\frac{v}{\lambda\sqrt{1+v^{2}/\lambda^{2}}}  \frac{u}{\lambda\sqrt{1+v^{2}/\lambda^{2}}}-\frac{uv^{2}}{\lambda^{3}\left(1+v^{2}/\lambda^{2}\right)^{\frac{3}{2}}}
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\end{array}\right|</math><math class="inline">=\frac{1}{\sqrt{1+v^{2}/\lambda^{2}}}\left[\frac{u}{\lambda\sqrt{1+v^{2}/\lambda^{2}}}-\frac{uv^{2}}{\lambda^{3}\left(1+v^{2}/\lambda^{2}\right)^{\frac{3}{2}}}\right]-\frac{-uv^{2}}{\lambda^{3}\left(1-v^{2}/\lambda^{2}\right)^{2}}</math><math class="inline">=\frac{u}{\lambda\left(1+v^{2}/\lambda^{2}\right)}=\frac{\lambda u}{\lambda^{2}+v^{2}}\qquad\left(\geq0\text{ because u is non-negative}\right).</math>
  
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<math>E\left[\mathbf{X}^{j}\mathbf{Y}^{k}\right]=\frac{\partial^{j+k}}{\partial\left(i\omega_{1}\right)^{j}\partial\left(i\omega_{2}\right)^{k}}\Phi_{\mathbf{XY}}\left(\omega_{1},\omega_{2}\right)|_{i\omega_{1}=i\omega_{2}=0}=\frac{\partial^{j+k}}{\partial\left(i\omega_{1}\right)^{j}\partial\left(i\omega_{2}\right)^{k}}\left[\left(1-i\omega_{1}\right)^{-1}\left(1-i\omega_{2}\right)^{-1}\right]|_{i\omega_{1}=i\omega_{2}=0}</math><math>=j!\left(1-i\omega_{1}\right)^{-\left(j+1\right)}k!\left(1-i\omega_{2}\right)^{-\left(k+1\right)}|_{i\omega_{1}=i\omega_{2}=0}=j!k!</math>  
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Because <math class="inline">\mathbf{X}</math> and <math class="inline">\mathbf{Y}</math> are statistically independent
  
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(e) Calculate the correlation coefficient <math>r_{\mathbf{XY}}</math>  between <math>\mathbf{X}</math>  and <math>\mathbf{Y}</math> .
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<math class="inline">f_{\mathbf{XY}}\left(x,y\right)=f_{\mathbf{X}}\left(x\right)f_{\mathbf{Y}}\left(y\right)=xe^{-x^{2}/2}\cdot\mathbf{1}_{\left[0,\infty\right)}\left(x\right)\cdot\frac{1}{\sqrt{2\pi}}e^{-y^{2}/2}=\frac{x}{\sqrt{2\pi}}e^{-\left(x^{2}+y^{2}\right)/2}\cdot\mathbf{1}_{\left[0,\infty\right)}\left(x\right).</math>  
  
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<math>r_{\mathbf{XY}}=\frac{Cov\left(\mathbf{X},\mathbf{Y}\right)}{\sigma_{\mathbf{X}}\sigma_{\mathbf{Y}}}=\frac{E\left[\mathbf{XY}\right]-E\left[\mathbf{X}\right]E\left[\mathbf{Y}\right]}{\sigma_{\mathbf{X}}\sigma_{\mathbf{Y}}}=\frac{1-1\cdot1}{\sigma_{\mathbf{X}}\sigma_{\mathbf{Y}}}=0.</math>
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Substituting these quantities, we get
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<math class="inline">f_{\mathbf{UV}}\left(u,v\right)=f_{\mathbf{XY}}\left(x\left(u,v\right),y\left(u,v\right)\right)\left|\frac{\partial\left(x,y\right)}{\partial\left(u,v\right)}\right|=\frac{u}{\sqrt{1+v^{2}/\lambda^{2}}}\cdot\frac{1}{\sqrt{2\pi}}e^{-u^{2}/2}\cdot\frac{\lambda u}{\lambda^{2}+v^{2}}\cdot\mathbf{1}_{\left[0,\infty\right)}\left(u\right)</math><math class="inline">=\frac{\lambda^{2}}{\sqrt{2\pi}}u^{2}e^{-u^{2}/2}\cdot\mathbf{1}_{\left[0,\infty\right)}\left(u\right)\cdot\frac{1}{\left(\lambda^{2}+v^{2}\right)^{\frac{3}{2}}}.</math>
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(b)
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Are <math class="inline">\mathbf{U}</math>  and <math class="inline">\mathbf{V}</math>  statistically independent? Justify your answer.
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<math class="inline">\mathbf{U}</math>  and <math class="inline">\mathbf{V}</math>  are statistically independent iff <math class="inline">f_{\mathbf{UV}}\left(u,v\right)=f_{\mathbf{U}}\left(u\right)f_{\mathbf{V}}\left(v\right)</math> .
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Now from part (a), we see that <math class="inline">f_{\mathbf{UV}}\left(u,v\right)=c_{1}g_{1}\left(u\right)\cdot c_{2}g_{2}\left(v\right)</math> where <math class="inline">g_{1}\left(u\right)=u^{2}e^{-u^{2}/2}\cdot\mathbf{1}_{\left[0,\infty\right)}\left(u\right)</math>  and <math class="inline">g_{2}\left(v\right)=\frac{1}{\left(\lambda^{2}+v^{2}\right)^{\frac{3}{2}}}</math>  with <math class="inline">c_{1}</math>  and <math class="inline">c_{2}</math>  selected such that <math class="inline">f_{\mathbf{U}}\left(u\right)=c_{1}g_{1}\left(u\right)</math>  and <math class="inline">f_{\mathbf{V}}\left(v\right)=c_{2}g_{2}\left(v\right)</math>  are both valid pdfs.
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<math class="inline">\therefore  \mathbf{U}</math> and <math class="inline">\mathbf{V}</math>  are statistically independent.
  
 
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Latest revision as of 12:10, 30 November 2010

Example. Two jointly distributed independent random variables

Let $ \mathbf{X} $ and $ \mathbf{Y} $ be two jointly distributed, independent random variables. The pdf of $ \mathbf{X} $ is

$ f_{\mathbf{X}}\left(x\right)=xe^{-x^{2}/2}\cdot\mathbf{1}_{\left[0,\infty\right)}\left(x\right) $, and $ \mathbf{Y} $ is a Gaussian random variable with mean 0 and variance 1 . Let $ \mathbf{U} $ and $ \mathbf{V} $ be two new random variables defined as $ \mathbf{U}=\sqrt{\mathbf{X}^{2}+\mathbf{Y}^{2}} $ and $ \mathbf{V}=\lambda\mathbf{Y}/\mathbf{X} $ where $ \lambda $ is a positive real number.

(a)

Find the joint pdf of $ \mathbf{U} $ and $ \mathbf{V} $ . (Direct pdf method)

$ f_{\mathbf{UV}}\left(u,v\right)=f_{\mathbf{XY}}\left(x\left(u,v\right),y\left(u,v\right)\right)\left|\frac{\partial\left(x,y\right)}{\partial\left(u,v\right)}\right| $

Solving for x and y in terms of u and v , we have $ u^{2}=x^{2}+y^{2} $ and $ v^{2}=\frac{\lambda^{2}y^{2}}{x^{2}}\Longrightarrow y^{2}=\frac{v^{2}x^{2}}{\lambda^{2}} $ .

Now, $ u^{2}=x^{2}+y^{2}=x^{2}+\frac{v^{2}x^{2}}{\lambda^{2}}=x^{2}\left(1+v^{2}/\lambda^{2}\right)\Longrightarrow x=\frac{u}{\sqrt{1+v^{2}/\lambda^{2}}}\Longrightarrow x\left(u,v\right)=\frac{u}{\sqrt{1+v^{2}/\lambda^{2}}} $ .

Thus, $ y=\frac{vx}{\lambda}=\frac{vu}{\lambda\sqrt{1+v^{2}/\lambda^{2}}}\Longrightarrow y\left(u,v\right)=\frac{vu}{\lambda\sqrt{1+v^{2}/\lambda^{2}}} $ .

Computing the Jacobian.

$ \frac{\partial\left(x,y\right)}{\partial\left(u,v\right)} =\left|\begin{array}{ll} \frac{\partial x}{\partial u} \frac{\partial x}{\partial v}\\ \frac{\partial y}{\partial u} \frac{\partial y}{\partial v} \end{array}\right|=\left|\begin{array}{cc} \frac{1}{\sqrt{1+v^{2}/\lambda^{2}}} \frac{-uv}{\lambda^{2}\left(1-v^{2}/\lambda^{2}\right)^{\frac{3}{2}}}\\ \frac{v}{\lambda\sqrt{1+v^{2}/\lambda^{2}}} \frac{u}{\lambda\sqrt{1+v^{2}/\lambda^{2}}}-\frac{uv^{2}}{\lambda^{3}\left(1+v^{2}/\lambda^{2}\right)^{\frac{3}{2}}} \end{array}\right| $$ =\frac{1}{\sqrt{1+v^{2}/\lambda^{2}}}\left[\frac{u}{\lambda\sqrt{1+v^{2}/\lambda^{2}}}-\frac{uv^{2}}{\lambda^{3}\left(1+v^{2}/\lambda^{2}\right)^{\frac{3}{2}}}\right]-\frac{-uv^{2}}{\lambda^{3}\left(1-v^{2}/\lambda^{2}\right)^{2}} $$ =\frac{u}{\lambda\left(1+v^{2}/\lambda^{2}\right)}=\frac{\lambda u}{\lambda^{2}+v^{2}}\qquad\left(\geq0\text{ because u is non-negative}\right). $

Because $ \mathbf{X} $ and $ \mathbf{Y} $ are statistically independent

$ f_{\mathbf{XY}}\left(x,y\right)=f_{\mathbf{X}}\left(x\right)f_{\mathbf{Y}}\left(y\right)=xe^{-x^{2}/2}\cdot\mathbf{1}_{\left[0,\infty\right)}\left(x\right)\cdot\frac{1}{\sqrt{2\pi}}e^{-y^{2}/2}=\frac{x}{\sqrt{2\pi}}e^{-\left(x^{2}+y^{2}\right)/2}\cdot\mathbf{1}_{\left[0,\infty\right)}\left(x\right). $

Substituting these quantities, we get

$ f_{\mathbf{UV}}\left(u,v\right)=f_{\mathbf{XY}}\left(x\left(u,v\right),y\left(u,v\right)\right)\left|\frac{\partial\left(x,y\right)}{\partial\left(u,v\right)}\right|=\frac{u}{\sqrt{1+v^{2}/\lambda^{2}}}\cdot\frac{1}{\sqrt{2\pi}}e^{-u^{2}/2}\cdot\frac{\lambda u}{\lambda^{2}+v^{2}}\cdot\mathbf{1}_{\left[0,\infty\right)}\left(u\right) $$ =\frac{\lambda^{2}}{\sqrt{2\pi}}u^{2}e^{-u^{2}/2}\cdot\mathbf{1}_{\left[0,\infty\right)}\left(u\right)\cdot\frac{1}{\left(\lambda^{2}+v^{2}\right)^{\frac{3}{2}}}. $

(b)

Are $ \mathbf{U} $ and $ \mathbf{V} $ statistically independent? Justify your answer.

$ \mathbf{U} $ and $ \mathbf{V} $ are statistically independent iff $ f_{\mathbf{UV}}\left(u,v\right)=f_{\mathbf{U}}\left(u\right)f_{\mathbf{V}}\left(v\right) $ .

Now from part (a), we see that $ f_{\mathbf{UV}}\left(u,v\right)=c_{1}g_{1}\left(u\right)\cdot c_{2}g_{2}\left(v\right) $ where $ g_{1}\left(u\right)=u^{2}e^{-u^{2}/2}\cdot\mathbf{1}_{\left[0,\infty\right)}\left(u\right) $ and $ g_{2}\left(v\right)=\frac{1}{\left(\lambda^{2}+v^{2}\right)^{\frac{3}{2}}} $ with $ c_{1} $ and $ c_{2} $ selected such that $ f_{\mathbf{U}}\left(u\right)=c_{1}g_{1}\left(u\right) $ and $ f_{\mathbf{V}}\left(v\right)=c_{2}g_{2}\left(v\right) $ are both valid pdfs.

$ \therefore \mathbf{U} $ and $ \mathbf{V} $ are statistically independent.

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