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Given a probability space $(\Omega, \mathcal{F}, P)$ consider a standard normal random variable $X$. Let $\tilde{X} = X + c$, $c \in \Bbb R$. Now consider the following probability measure $\tilde{P}(A) = \int_A \frac{e^{-\frac{(X+c)^2}{2}}}{e^{-\frac{X^2}{2}}} dP$. It is easy to check that with respect to $\tilde{P}$, $\tilde{X}$ is a standard normal random variable.

I want to know the intuition behind constructing such $\tilde{P}$. I mean why the integrand should be in that form ?

Thanks.

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  • $\begingroup$ I don't think there is much of an intuition needed here. The idea is just to shift the mean in the distribution of $\bar{X}$ by changing the probability measure with something that cancels out the non-zero mean in that distribution. Since the mean of a normal r.v. appears in an exponential, it is no surprise that the measure-changing variable (Radon Nikodym derivative) has an exponential term in it as well. $\endgroup$ – Calculon Jul 29 '15 at 15:31
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First, assuming $A \in \sigma(X)$ and $B \in \sigma(\tilde{X})$, your statement $$ \tilde{P}(A) = \int_A \frac{e^{-\frac{(X+c)^2}{2}}}{e^{-\frac{X^2}{2}}} dP $$ should be $$ \tilde{P}(A) = \int_A e^{-Xc - \frac{1}{2}c^2} dP. \qquad (1) $$ Then $$ \tilde{P}(A) = \frac{1}{\sqrt{2\pi}}\int_A e^{-xc - \frac{1}{2}c^2} e^{-\frac{1}{2}x^2} dx = \frac{1}{\sqrt{2\pi}}\int_{B} e^{-\frac{1}{2}\tilde{x}^2} d\tilde{x} $$ as desired.


To see why this is the integrand needed, more generally let $X \sim \mathcal{N}(\mu, \sigma^2)$. Then $Y := X + c \sim \mathcal{N}(\mu + c, \sigma^2)$. The probability density function (pdf) of $Y$ is $$ f_Y(y) = \frac{1}{\sigma\sqrt{2\pi}}\exp\left(-\frac{1}{2}\left(\frac{y - (\mu + c)}{\sigma}\right)^2\right), $$ and so $P(Y \in B) = \int_B f_Y(y) \, dy$. So, how can we modify $f_Y$ such that when we compute probabilities using this modified density, $Y \sim \mathcal{N}(\mu,\sigma^2)$? I.e., we want $$ \tilde{P}(Y \in B) = \int_B \frac{1}{\sigma\sqrt{2\pi}}\exp\left(-\frac{1}{2}\left(\frac{y - \mu}{\sigma}\right)^2\right) dy. $$ Let's see, working with just the exponential argument: \begin{align*} -\frac{1}{2\sigma^2}\left(y - (\mu + c)\right)^2 & = -\frac{1}{2\sigma^2}\left(y^2 -2y(\mu + c) + (\mu +c)^2\right) \\ & = -\frac{1}{2\sigma^2}\left((y^2 -2y\mu + \mu^2) -2yc + 2\mu c + c^2\right) \\ & = -\frac{1}{2}\left(\frac{y - \mu}{\sigma}\right)^2 -\frac{1}{2\sigma^2}\left(-2yc + 2\mu c + c^2\right). \end{align*} So we multiply $f_Y(y)$ by $\exp\left(\frac{1}{2\sigma^2}\left(-2yc + 2\mu c + c^2\right)\right)$.

Put another way, the Radon-Nikodym derivative of $\tilde{P}$ w.r.t. $P$ is $\exp\left(\frac{1}{2\sigma^2}\left(-2yc + 2\mu c + c^2\right)\right)$. In your case with $\mu = 0$ and $\sigma = 1$, $\frac{d \tilde{P}}{d P}$ = $\exp\left(\frac{1}{2}\left(-2yc + c^2\right)\right) = \exp(-xc -\frac{1}{2}c^2)$, hence (1).


This technique is commonly called "exponential twisting." For a normal random variable $X \sim \mathcal{N}(\mu,\sigma^2)$, define the pdf $$ f_t(x) = \frac{e^{tx}f_X(x)}{M_X(t)} $$ where $M_X$ is the moment-generating function (MGF) for $X$. Then you can easily show that $f_t$ is the pdf for the $\mathcal{N}(\mu + \sigma^2t, \sigma^2)$ distribution.

Try it: Replace $X$ here with $Y$ above, replace $f_X$ with $f_Y$ above and compute the MGF for $Y$. It should be clear we need to set $t = -\frac{c}{\sigma^2}$, and you'll see that $\frac{e^{-\frac{c}{\sigma^2}y}}{M_Y(-\frac{c}{\sigma^2})} = \frac{d \tilde{P}}{d P}$ as above.

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  • $\begingroup$ Thanks for the explanation. $\endgroup$ – Sosha Jul 30 '15 at 6:08

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