Simplifying a Random-Walk Identity Using the Probability Constraint

Summary

The document checks a logarithmic identity appearing in a solution to a random-walk problem. The stated probabilities satisfy q = 1 − p, with p below one half, so the expression under the square root becomes the square of 1 − 2p. Because this quantity is positive in the stated range, the positive square root equals 1 − 2p.

Substituting that result into the fraction reduces it to (1 − p)/p, which establishes the equality after taking logarithms. The explanation is a short algebraic verification rather than a trading method or empirical study. Its conclusion relies on the given probability constraint and range; outside those assumptions, the square-root simplification may require a different sign or may not apply.

Key ideas

• The constraint q = 1 − p turns the radicand into (1 − 2p) squared.
• Since p is below one half, the positive square root is 1 − 2p.
• Substituting the root simplifies the fraction to (1 − p)/p.
• The logarithmic identity follows under the stated probability conditions.

Tags

# Identity given in Shreve volume 1


# Identity given in Shreve volume 1












in a solution to a question about random walks (5.3 i), Part of the answer includes the identity:

$$\ln \frac{1+\sqrt{1-4 pq}}{2p}=\ln\frac{1-p}{p}$$

note that $p+q=1$ and $0<p<1/2<q<1$.

this does not seem true to me, even when I restrict the values of $p,q$ appropriately. What am I missing?

## Answer by Olaf (score 3, accepted)

https://quant.stackexchange.com/a/17347

Using $q = 1-p$ we can work out the root as: $$\sqrt{1-4pq} = \sqrt{1-4p(1-p)} = \sqrt{1-4p+4p^2} = \sqrt{(1-2p)^2}$$

Taking the positive root reduces this to $(1-2p)$. This gives for the fraction:

$$\frac{1 + \sqrt{1-4pq}}{2p} = \frac{1 + (1-2p)}{2p} = \frac{1-p}{p}$$

This also holds inside the logarithm.

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.