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=Example. Two jointly distributed independent random variables= | =Example. Two jointly distributed independent random variables= | ||
| − | Let <math>\mathbf{X}</math> and <math>\mathbf{Y}</math> be two jointly distributed, independent random variables. The pdf of <math>\mathbf{X}</math> is | + | 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 |
| − | <math>f_{\mathbf{X}}\left(x\right)=xe^{-x^{2}/2}\cdot\mathbf{1}_{\left[0,\infty\right)}\left(x\right)</math>, and <math>\mathbf{Y}</math> is a Gaussian random variable with mean 0 and variance 1 . Let <math>\mathbf{U}</math> and <math>\mathbf{V}</math> be two new random variables defined as <math>\mathbf{U}=\sqrt{\mathbf{X}^{2}+\mathbf{Y}^{2}}</math> and <math>\mathbf{V}=\lambda\mathbf{Y}/\mathbf{X}</math> where <math>\lambda</math> is a positive real number. | + | <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. |
(a) | (a) | ||
| − | Find the joint pdf of <math>\mathbf{U}</math> and <math>\mathbf{V}</math> . (Direct pdf method) | + | Find the joint pdf of <math class="inline">\mathbf{U}</math> and <math class="inline">\mathbf{V}</math> . (Direct pdf method) |
| − | <math>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> | + | <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> |
| − | Solving for x and y in terms of u and v , we have <math>u^{2}=x^{2}+y^{2}</math> and <math>v^{2}=\frac{\lambda^{2}y^{2}}{x^{2}}\Longrightarrow y^{2}=\frac{v^{2}x^{2}}{\lambda^{2}}</math> . | + | 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> . |
| − | Now, <math>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> . | + | 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> . |
| − | Thus, <math>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> . | + | 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> . |
Computing the Jacobian. | Computing the Jacobian. | ||
| − | <math>\frac{\partial\left(x,y\right)}{\partial\left(u,v\right)} =\left|\begin{array}{ll} | + | <math class="inline">\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 x}{\partial u} \frac{\partial x}{\partial v}\\ | ||
\frac{\partial y}{\partial u} \frac{\partial y}{\partial v} | \frac{\partial y}{\partial u} \frac{\partial y}{\partial v} | ||
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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}}}\\ | \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}}} | \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|</math><math>=\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>=\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> | + | \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> |
| − | Because <math>\mathbf{X}</math> and <math>\mathbf{Y}</math> are statistically independent | + | Because <math class="inline">\mathbf{X}</math> and <math class="inline">\mathbf{Y}</math> are statistically independent |
| − | <math>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> | + | <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> |
Substituting these quantities, we get | Substituting these quantities, we get | ||
| − | <math>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>=\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> | + | <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> |
(b) | (b) | ||
| − | Are <math>\mathbf{U}</math> and <math>\mathbf{V}</math> statistically independent? Justify your answer. | + | Are <math class="inline">\mathbf{U}</math> and <math class="inline">\mathbf{V}</math> statistically independent? Justify your answer. |
| − | <math>\mathbf{U}</math> and <math>\mathbf{V}</math> are statistically independent iff <math>f_{\mathbf{UV}}\left(u,v\right)=f_{\mathbf{U}}\left(u\right)f_{\mathbf{V}}\left(v\right)</math> . | + | <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> . |
| − | Now from part (a), we see that <math>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>g_{1}\left(u\right)=u^{2}e^{-u^{2}/2}\cdot\mathbf{1}_{\left[0,\infty\right)}\left(u\right)</math> and <math>g_{2}\left(v\right)=\frac{1}{\left(\lambda^{2}+v^{2}\right)^{\frac{3}{2}}}</math> with <math>c_{1}</math> and <math>c_{2}</math> selected such that <math>f_{\mathbf{U}}\left(u\right)=c_{1}g_{1}\left(u\right)</math> and <math>f_{\mathbf{V}}\left(v\right)=c_{2}g_{2}\left(v\right)</math> are both valid pdfs. | + | 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. |
| − | <math>\therefore \mathbf{U}</math> and <math>\mathbf{V}</math> are statistically independent. | + | <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 12:08, 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.
