Bernoulli's Inequality/Corollary/General Result
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Theorem
For all $n \in \Z_{\ge 0}$:
- $\ds \prod_{j \mathop = 1}^n \paren {1 - a_j} \ge 1 - \sum_{j \mathop = 1}^n a^j$
where $0 < a_j < 1$ for all $j$.
Proof
The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
- $\ds \prod_{j \mathop = 1}^n \paren {1 - a_j} \ge 1 - \sum_{j \mathop = 1}^n a^j$
$\map P 0$ is the case:
\(\ds \prod_{j \mathop = 1}^0 \paren {1 - a_j}\) | \(=\) | \(\ds 1\) | Definition of Vacuous Product | |||||||||||
\(\ds \) | \(\ge\) | \(\ds 1 - 0\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1 - \sum_{j \mathop = 1}^0 a^j\) | Definition of Vacuous Summation |
Thus $\map P 0$ is seen to hold.
Basis for the Induction
$\map P 1$ is the case:
\(\ds \prod_{j \mathop = 1}^1 \paren {1 - a_j}\) | \(=\) | \(\ds 1 - a_j\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1 - \sum_{j \mathop = 1}^1 a^j\) |
Thus $\map P 1$ is seen to hold.
This is the basis for the induction.
Induction Hypothesis
Now it needs to be shown that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
- $\ds \prod_{j \mathop = 1}^k \paren {1 - a_j} \ge 1 - \sum_{j \mathop = 1}^k a^j$
from which it is to be shown that:
- $\ds \prod_{j \mathop = 1}^{k + 1} \paren {1 - a_j} \ge 1 - \sum_{j \mathop = 1}^{k + 1} a^j$
Induction Step
This is the induction step:
\(\ds \prod_{j \mathop = 1}^{k + 1} \paren {1 - a_j}\) | \(=\) | \(\ds \paren {\prod_{j \mathop = 1}^k \paren {1 - a_j} } \paren {1 - a_{k + 1} }\) | Definition of Continued Product | |||||||||||
\(\ds \) | \(\ge\) | \(\ds \paren {1 - \sum_{j \mathop = 1}^k a^j} \paren {1 - a_{k + 1} }\) | Induction Hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds 1 - \sum_{j \mathop = 1}^k a^j - a_{k + 1} + a_{k + 1} \sum_{j \mathop = 1}^k a^j\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1 - \sum_{j \mathop = 1}^{k + 1} a^j + a_{k + 1} \sum_{j \mathop = 1}^k a^j\) | ||||||||||||
\(\ds \) | \(\ge\) | \(\ds 1 - \sum_{j \mathop = 1}^{k + 1} a^j\) |
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
- $\ds \forall n \in \Z_{\ge 0}: \prod_{j \mathop = 1}^n \paren {1 - a_j} \ge 1 - \sum_{j \mathop = 1}^n a^j$
$\blacksquare$
Sources
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.3$: Sums and Products: Exercise $27$