## 9.1.4 Conditional Expectation (MMSE)

Remember that the posterior distribution, $f_{X|Y}(x|y)$, contains all the knowledge that we have about the unknown quantity $X$. Therefore, to find a point estimate of $X$, we can just choose a summary statistic of the posterior such as its mean, median, or mode. If we choose the mode (the value of $x$ that maximizes $f_{X|Y}(x|y)$), we obtain the MAP estimate of $X$. Another option would be to choose the posterior mean, i.e., \begin{align} \hat{x}=E[X|Y=y]. \end{align} We will show that $E[X|Y=y]$ will give us the best estimate of $X$ in terms of the mean squared error. For this reason, the conditional expectation is called the minimum mean squared error (MMSE) estimate of $X$. It is also called the least mean squares (LMS) estimate or simply the Bayes' estimate of $X$.
Minimum Mean Squared Error (MMSE) Estimation

The minimum mean squared error (MMSE) estimate of the random variable $X$, given that we have observed $Y=y$, is given by \begin{align} \hat{x}_{M}=E[X|Y=y]. \end{align}

Example
Let $X$ be a continuous random variable with the following PDF \begin{align} \nonumber f_X(x) = \left\{ \begin{array}{l l} 2x & \quad \textrm{if }0 \leq x \leq 1 \\ & \quad \\ 0 & \quad \text{otherwise} \end{array} \right. \end{align} We also know that \begin{align} \nonumber f_{Y|X}(y|x) = \left\{ \begin{array}{l l} 2xy-x+1 & \quad \textrm{if }0 \leq y \leq 1 \\ & \quad \\ 0 & \quad \text{otherwise} \end{array} \right. \end{align} Find the MMSE estimate of $X$, given $Y=y$ is observed.
• Solution
• First we need to find the posterior density, $f_{X|Y}(x|y)$. We have \begin{align} f_{X|Y}(x|y)=\frac{f_{Y|X}(y|x)f_{X}(x)}{f_{Y}(y)}. \end{align} We can find $f_{Y}(y)$ as \begin{align} f_Y(y)&=\int_{0}^{1} f_{Y|X}(y|x)f_{X}(x) dx\\ &=\int_{0}^{1} (2xy-x+1) 2x dx\\ &=\frac{4}{3}y+\frac{1}{3}, \quad \textrm{ for }0 \leq y \leq 1. \end{align} We conclude \begin{align} f_{X|Y}(x|y)=\frac{6x(2xy-x+1)}{4y+1}, \quad \textrm{ for }0 \leq x \leq 1. \end{align} The MMSE estimate of $X$ given $Y=y$ is then given by \begin{align} \hat{x}_{M}&=E[X|Y=y]\\ &=\int_{0}^{1} x f_{X|Y}(x|y) dx \\ &=\frac{1}{4y+1} \int_{0}^{1} 6x^2(2xy-x+1) dx\\ &=\frac{3y+ \frac{1}{2}}{4y+1}. \end{align}

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