CF 103828E - Do you where is Naseem?
Let $G(z)=sum{x1=0}^{1}cdotssum{xn=0}^{1} z^{x1+cdots+xn} f(x1,ldots,xn)$ be the generating function defined in Exercise 25, and let $F(p)$ denote the reliability polynomial when $p1=cdots=pn=p$, so that $$F(p)=sum{x1=0}^{1}cdotssum{xn=0}^{1} (1-p)^{1-x1}p^{x1}cdots…
CF 103828E - Do you where is Naseem?
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Solution
Solution
Let $G(z)=\sum_{x_1=0}^{1}\cdots\sum_{x_n=0}^{1} z^{x_1+\cdots+x_n} f(x_1,\ldots,x_n)$ be the generating function defined in Exercise 25, and let $F(p)$ denote the reliability polynomial when $p_1=\cdots=p_n=p$, so that
$$F(p)=\sum_{x_1=0}^{1}\cdots\sum_{x_n=0}^{1} (1-p)^{1-x_1}p^{x_1}\cdots (1-p)^{1-x_n}p^{x_n} f(x_1,\ldots,x_n).$$
For a fixed vector $x=(x_1,\ldots,x_n)$, the product factors according to the Hamming weight $w(x)=x_1+\cdots+x_n$, giving
$$(1-p)^{1-x_1}p^{x_1}\cdots (1-p)^{1-x_n}p^{x_n}=(1-p)^{n-w(x)}p^{w(x)}.$$
Substitution yields
$$F(p)=\sum_{x} f(x), (1-p)^{n-w(x)} p^{w(x)}.$$
Factoring out $(1-p)^n$ produces
$$F(p)=(1-p)^n \sum_{x} f(x)\left(\frac{p}{1-p}\right)^{w(x)}.$$
The remaining sum matches the definition of $G(z)$ evaluated at $z=\frac{p}{1-p}$, since $z^{w(x)}$ appears with the same exponent structure. Therefore,
$$F(p)=(1-p)^n G!\left(\frac{p}{1-p}\right).$$
This expresses $F(p)$ as a simple rescaling of the generating function, obtained by a single substitution followed by multiplication by a monomial factor. This completes the solution. ∎