Revision as of 12:06, 30 November 2010

Example. Two jointly distributed random variables

Two joinly distributed random variables $ \mathbf{X} $ and $ \mathbf{Y} $ have joint pdf

$ f_{\mathbf{XY}}\left(x,y\right)=\begin{cases} \begin{array}{ll} c ,\text{ for }x\geq0,y\geq0,\textrm{ and }x+y\leq1\\ 0 ,\text{ elsewhere.} \end{array}\end{cases} $

(a)

Find the constant c such that $ f_{\mathbf{XY}}(x,y) $ is a valid pdf.

File:002.eps

$ \iint_{\mathbf{R}^{2}}f_{\mathbf{XY}}\left(x,y\right)=c\cdot \text{Area}=1 $ where $ \text{Area}=\frac{1}{2}. $

$ \therefore c=2 $

(b)

Find the conditional density of $ \mathbf{Y} $ conditioned on $ \mathbf{X}=x $ .

$ f_{\mathbf{Y}}\left(y|\left\{ \mathbf{X}=x\right\} \right)=\frac{f_{\mathbf{XY}}\left(x,y\right)}{f_{\mathbf{X}}(x)}. $

$ f_{\mathbf{X}}(x)=\int_{-\infty}^{\infty}f_{\mathbf{XY}}\left(x,y\right)dy=\int_{0}^{1-x}2dy=2\left(1-x\right)\cdot\mathbf{1}_{\left[0,1\right]}(x). $

$ f_{\mathbf{Y}}\left(y|\left\{ \mathbf{X}=x\right\} \right)=\frac{f_{\mathbf{XY}}\left(x,y\right)}{f_{\mathbf{X}}(x)}=\frac{2}{2\left(1-x\right)}=\frac{1}{1-x}\textrm{ where }0\leq y\leq1-x\Longrightarrow\frac{1}{1-x}\cdot\mathbf{1}_{\left[0,1-x\right]}\left(y\right). $

(c)

Find the minimum mean-square error estimator $ \hat{y}_{MMS}\left(x\right) $ of $ \mathbf{Y} $ given that $ \mathbf{X}=x $ .

$ \hat{y}_{MMS}\left(x\right)=E\left[\mathbf{Y}|\left\{ \mathbf{X}=x\right\} \right]=\int_{\mathbf{R}}yf_{\mathbf{Y}}\left(y|\left\{ \mathbf{X}=x\right\} \right)dy=\int_{0}^{1-x}\frac{y}{1-x}dy=\frac{y^{2}}{2\left(1-x\right)}\biggl|_{0}^{1-x}=\frac{1-x}{2}. $

(d)

Find a maximum aposteriori probability estimator.

$ \hat{y}_{MAP}\left(x\right)=\arg\max_{y}\left\{ f_{Y}\left(y|\left\{ \mathbf{X}=x\right\} \right)\right\} $ but $ f_{Y}\left(y|\left\{ \mathbf{X}=x\right\} \right)=\frac{1}{1-x}\cdot\mathbf{1}_{\left[0,1-x\right]}\left(y\right) $ . Any $ \hat{y}\in\left[0,1-x\right] $ is a MAP estimator. The MAP estimator is NOT unique.

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Example. Two jointly distributed random variables

Two joinly distributed random variables $ \mathbf{X} $ and $ \mathbf{Y} $ have joint pdf

$ f_{\mathbf{XY}}\left(x,y\right)=\begin{cases} \begin{array}{ll} c ,\text{ for }x\geq0,y\geq0,\textrm{ and }x+y\leq1\\ 0 ,\text{ elsewhere.} \end{array}\end{cases} $

(a)

Find the constant c such that $ f_{\mathbf{XY}}(x,y) $ is a valid pdf.

File:002.eps

$ \iint_{\mathbf{R}^{2}}f_{\mathbf{XY}}\left(x,y\right)=c\cdot Area=1 $ where $ Area=\frac{1}{2} $ .

$ \therefore c=2 $

(b)

Find the conditional density of $ \mathbf{Y} $ conditioned on $ \mathbf{X}=x $ .

$ f_{\mathbf{Y}}\left(y|\left\{ \mathbf{X}=x\right\} \right)=\frac{f_{\mathbf{XY}}\left(x,y\right)}{f_{\mathbf{X}}(x)}. $

$ f_{\mathbf{X}}(x)=\int_{-\infty}^{\infty}f_{\mathbf{XY}}\left(x,y\right)dy=\int_{0}^{1-x}2dy=2\left(1-x\right)\cdot\mathbf{1}_{\left[0,1\right]}(x). $

$ f_{\mathbf{Y}}\left(y|\left\{ \mathbf{X}=x\right\} \right)=\frac{f_{\mathbf{XY}}\left(x,y\right)}{f_{\mathbf{X}}(x)}=\frac{2}{2\left(1-x\right)}=\frac{1}{1-x}\textrm{ where }0\leq y\leq1-x\Longrightarrow\frac{1}{1-x}\cdot\mathbf{1}_{\left[0,1-x\right]}\left(y\right). $

(c)

Find the minimum mean-square error estimator $ \hat{y}_{MMS}\left(x\right) $ of $ \mathbf{Y} $ given that $ \mathbf{X}=x $ .

$ \hat{y}_{MMS}\left(x\right)=E\left[\mathbf{Y}|\left\{ \mathbf{X}=x\right\} \right]=\int_{\mathbf{R}}yf_{\mathbf{Y}}\left(y|\left\{ \mathbf{X}=x\right\} \right)dy=\int_{0}^{1-x}\frac{y}{1-x}dy=\frac{y^{2}}{2\left(1-x\right)}\biggl|_{0}^{1-x}=\frac{1-x}{2}. $

(d)

Find a maximum aposteriori probability estimator.

$ \hat{y}_{MAP}\left(x\right)=\arg\max_{y}\left\{ f_{Y}\left(y|\left\{ \mathbf{X}=x\right\} \right)\right\} $ but $ f_{Y}\left(y|\left\{ \mathbf{X}=x\right\} \right)=\frac{1}{1-x}\cdot\mathbf{1}_{\left[0,1-x\right]}\left(y\right) $ . Any $ \hat{y}\in\left[0,1-x\right] $ is a MAP estimator. The MAP estimator is NOT unique.

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