Changing Probability Measures to Make a Double-Exponential Variable Normal

Summary

The answer shows how to construct a new probability measure under which a random variable with a valid density becomes standard normal. It defines the new measure using a Radon–Nikodym derivative equal to the standard normal density divided by the variable’s original density, evaluated at that variable. Integrating under the original measure then cancels the original density and recovers the standard normal cumulative distribution function.

The reciprocal density ratio gives the reverse change of measure. This is a mathematical measure-change argument that can be useful in probability and quantitative finance, but the excerpt does not provide the original double-exponential density or discuss conditions beyond its positivity and normalization. In particular, practical use requires the relevant measures to be well defined and mutually absolutely continuous on the range considered.

Key ideas

• A positive density that integrates to one defines a valid probability distribution.
• A density ratio can define a new measure through a Radon–Nikodym derivative.
• Choosing the standard normal density in the ratio makes the variable standard normal under the new measure.
• The reciprocal density ratio expresses the change back to the original measure.

Tags

# Please help me with this problem of double exponential distribution


# Please help me with this problem of double exponential distribution












please help me with this problem of double exponential distribution

## Answer by Gordon (score 1, accepted)

https://quant.stackexchange.com/a/22347

Since $f_X(x) > 0$ and \begin{align*} \int_{-\infty}^{\infty} f_X(x) dx = 1, \end{align*} $f_X(x)$ is a valid density function.

Let \begin{align*} \varphi(x) =\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}} \end{align*} be the density function of a standard normal random variable. We define the measure $\widetilde{P}$ using the Randon-Nykodim derivative \begin{align*} \frac{d\widetilde{P}}{dP} = \frac{\varphi(X)}{f_X(X)}. \end{align*} Then, \begin{align*} \widetilde{P}(X \leq x) &= E_{\widetilde{P}}(1_{X \le x})\\ &=E_P\left(\frac{d\widetilde{P}}{dP} 1_{X \le x} \right)\\ &=E_P\left(\frac{\varphi(X)}{f_X(X)} 1_{X \le x} \right)\\ &=\int_{-\infty}^x \frac{\varphi(x)}{f_X(x)} f_X(x) dx\\ &=\int_{-\infty}^x \varphi(x) dx. \end{align*} That is, $X$ is standard normal w.r.t. the measure $\widetilde{P}$.

The equivalence follows, since \begin{align*} \frac{dP}{d\widetilde{P}} = \frac{f_X(X)}{\varphi(X)}. \end{align*}

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

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