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Math and science::INF ML AI
One difference of squares from two. Posteriors with Gaussians.
\( (a-x)^2 + (x-b)^2 \)
This expression appears when dealing with posterior probabilities for the
mean parameter of a Gaussian distribution. [Can you remember how?]
Minimum at \( x = \frac{a + b}{2} \)
Let \( a < b \) and define \( d=b - a \). If \( x \) is constrained to be between \( a \) and \( b \), then the
expression is maximized when \( x=a \) or lx=b \). The maximum value is \( d^2 \). If \( x \) is not constrained to
\( [a, b] \), then the value of the expression can grow unbounded. The expression is minimized when \( x=\frac{d}{2}
\). Let \( h=\frac{d}{2} \) be the half-width of the interval. When \( x \) is at the midpoint, \( x=a + h \), then
the expression becomes:
\[
h^2 + h^2 = 2h^2 = \frac{d^2}{2}
\]
If you increase \( x \) slightly, \( x \to x + \varepsilon \), then the expression would become:
\[
(h + \varepsilon)^2 + (h - \varepsilon)^2 = 2h^2 + 2\varepsilon^2.
\]
The expression increases in value. It was indeed minimized at the midpoint when \( x = a + h \), taking the value \( 2h^2 = \frac{d^2}{2} \).
From two squares to one square
The expression:
\[
(a-x)^2 + (x-b)^2
\]
can be rewritten as one square involving \( x \):
\[
\begin{aligned}
(a-x)^2 + (x-b)^2
&= a^2 + x^2 - 2ax + x^2 + b^2 - 2bx \\
&= 2x^2 - 2x(a + b) + a^2 + b^2 \\
&= 2((x - \frac{a + b}{2})^2 - \frac{(a + b)^2}{4}) + a^2 + b^2 \\
&= 2(x - \frac{a + b}{2})^2 - \frac{(a^2 + 2ab + b^2)}{2} + \frac{2a^2 + 2b^2}{2} \\
&= 2(x - \frac{a + b}{2})^2 + \frac{a^2 + b^2 - 2ab}{2} \\
&= 2(x - \frac{a + b}{2})^2 + \frac{(a - b)^2}{2}
\end{aligned}
\]
Here again we see that the expression is minimized when \( x = \frac{a +
b}{2} \), and the minimum value is \( \frac{(a - b)^2}{2} \).
