ⓘ Zero bias transform
The zerobias transform is a transform from one probability distribution to another. The transform arises in applications of Steins method in probability and statistics.
1. Formal definition
The zero bias transform may be applied to both discrete and continuous random variables. The zero bias transform of a density function f t, defined for all real numbers t ≥ 0, is the function g s, defined by
g s = ∫ s ∞ t f t 1 t > s d t {\displaystyle gs=\int _{s}^{\infty }tft1t> s\,dt}where s and t are real numbers and f t is the density or mass function of the random variable T.
An equivalent but alternative approach is to deduce the nature of the transformed random variable by evaluating the expected value
E T H T) = σ 2 E h T z) {\displaystyle \operatorname {E} THT)=\sigma ^{2}EhT^{z})}where the rightside superscript denotes a zero biased random variable whereas the left hand side expectation represents the original random variable. An example from each approach is given in the examples section beneath.
If the random variable is discrete the integral becomes a sum from positive infinity to s. The zero bias transform is taken for a mean zero, variance 1 random variable which may require a locationscale transform to the random variable.
2. Applications
The zero bias transformation arises in applications where a normal approximation is desired. Similar to Steins method the zero bias transform is often applied to sums of random variables with each summand having finite variance an mean zero.
The zero bias transform has been applied to CDO tranche pricing.
3. Examples
1. Consider a Bernoullip random variable B with PrB = 0 = 1 − p. The zero bias transform of T = B − p is:
E T H T) = − p 1 − p H − p + 1 − p H 1 − p = p 1 − p \\&=p1p\int _{p}^{1p}hs\,ds\end{aligned}}}where h is the derivative of H. From there it follows that the random variable is a continuous uniform random variable on the support − p, 1 − p. This example shows how the zero bias transform smooths a discrete distribution into a continuous distribution.
2. Consider the continuous uniform on the support − 3, 3 {\displaystyle {\sqrt {3}},{\sqrt {3}}}.
∫ s 3 t 1 t > s f t d t = ∫ s 3 t 2 3 d t = 3 4 − s 2 3 4 where − 3 < s < 3 {\displaystyle \int _{s}^{\sqrt {3}}t1t> sft\,dt=\int _{s}^{\sqrt {3}}{\frac {t}{2{\sqrt {3}}}}\,dt={\frac {\sqrt {3}}{4}}{\frac {s^{2}} positive voltage at the cathode of the photodiode applies a reverse bias This reverse bias increases the width of the depletion region and lowers the junction
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