Q.7.74 - A First Course in Probability

Question

In Example 6c, suppose that $X$ is uniformly distributed over $(0,1)$. If the discretized regions are determined by $a_0=0,a_1=\frac{1}{2}$ and $a_2=1$. calculate the optimal quantizer $Y$ and compute $E\left[{(X-Y)}^2\right].$

Step-by-Step Solution

Answer

The optimal quantizer $Y$ is $\frac{1}{2}$. And the computation of $E\left[{(X-Y)}^2\right]$ is $\frac{1}{12}.$

Step 1: Given Information

$X=$Uniformly distributed over$(0,1)$

$a_0=0,$

$a_1=\frac{1}{2}$

And $a_2=1$

Step 2: Explanation

If $\mathrm{X}~\mathrm{U}(0,1)\Rightarrow {\mathrm{F}}_{\mathrm{X}}\left(x\right)=x;0<x<1$

$\therefore Y=\left\{\begin{array}{l}{y}_o\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{ if }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0<x\le \frac{1}{2}\\ y_1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{ if }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\frac{1}{2}<x\le 1\end{array}\right.$

$\begin{array}{r}P\left[Y=y_0\right]=F_X\left(\frac{1}{2}\right)-F_X\left(0\right)\end{array}$

$\begin{array}{r}={\int }_0^{\frac{1}{2}} dx-{\int }_0^0 dx=\frac{1}{2}\end{array}$

$P\left[Y=y_1\right]={\int }_0^1 dx-{\int }_0^{\frac{1}{2}} dx$

$=1-\frac{1}{2}=\frac{1}{2}$

$\therefore \mathrm{Var}\left(Y\right)=0\left(\because \mathrm{P}\left[\mathrm{Y}=y_0\right]=P\left[Y=y_1\right]=\frac{1}{2}\right)$

Step 3: Explanation

Now the quantity is minimized when,

$\begin{array}{r}{y}_0=\mathrm{E}[\mathrm{X}\mid \mathrm{I}=0]\end{array}$

style="max-width: none; vertical-align: -28px;" $\begin{array}{r}={\int }_0^{\frac{1}{2}} \frac{x}{\frac{1}{2}}dx\end{array}$

$={\int }_0^{\frac{1}{2}} 2xdx$

$=\frac{1}{4}$

$\begin{array}{r}{y}_1=\mathrm{E}[\mathrm{X}\mid \mathrm{I}=1]\end{array}$

$={\int }_{\frac{1}{2}}^1 \frac{x}{2}dx$

$={\int }_{\frac{1}{2}}^1 2xdx=\frac{3}{4}$

Step 4: Explanation

Distribution of $Y$ is

$\mathrm{P}\left[\mathrm{Y}=\frac{1}{4}\right]=\mathrm{P}\left[\mathrm{Y}=\frac{3}{4}\right]=\frac{1}{2}$

And $Var\left(Y\right)=0$

And $Var\left(X\right)=\frac{4-3}{12}$

$=\frac{1}{12}$

$\begin{array}{r}\therefore \mathrm{E}\left[(\mathrm{X}-\mathrm{Y}{)}^2\right]=\mathrm{Var}\left(\mathrm{X}\right)-\mathrm{Var}\left(\mathrm{Y}\right)\end{array}$

$=\frac{1}{12}-0$

$=\frac{1}{12}$

Step 5: Final Answer

Therefore, the optimal quantizer $Y$ is $\frac{1}{2}$. And the computation of $E\left[(X-Y{)}^2\right]$ is $\frac{1}{12}.$