Computing transformations of probability distributions

Question

Are there any good techniques for solving the following sort of problem. Assume that $X\sim N(\mu,\sigma^2+\eta^2)$. Find a function $\varphi:\mathbb{R}\to\mathbb{R}$, such that $\varphi(X)\sim N(\mu,\sigma^2)$? To make the problem simpler, assume we want to compute $\phi=\varphi^{-1}$ instead.

My approach has been that if we choose $Y\sim N(\mu,\sigma^2)$, then we get $$f_Y(y) = f_X(\phi(y))\frac{d\phi}{dy}(y),$$ but the equation seems to become a bit tricky to handle. Are there any better techniques or can someone see how to solve the resulting equation?

Answer 1

A promising transformation is

$$Y=\phi(X)=aX+b.$$

The question is: how to choose $a$ and $b$ so that $\sigma_Y^2=\sigma^2$ and $\mu_Y=\mu.$

Let's calculate the new variance first $$\sigma_Y^2=E[(aX+b)^2]-E^2[aX+b]=$$ $$=a^2E[X^2]+2ab\mu+b^2-a^2\mu^2-2a\mu b-b^2=a^2\left(E[X^2]-\mu^2\right)=a^2(\sigma^2+\eta^2)\tag 1$$

Then, the new expectation

$$\mu_Y=E[aX+b]=a\mu+b.\tag 2$$

We want that the new variance equal $\sigma^2$. From $(1)$ we have to solve the following equation:

$$a^2(\sigma^2+\eta^2)=\sigma^2.$$

So,

$$a^2=\frac{\sigma^2}{\sigma^2+\eta^2}.$$

And we want to keep the original $\mu.$ That is, from $(2)$ and $(1)$:

$$\mu=a\mu+b=\frac{\sigma}{\sqrt{\sigma^2+\eta^2}}\mu+b.$$

So,

$$b=\mu\left(1-\frac{\sigma}{\sqrt{\sigma^2+\eta^2}}\right).$$

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The solution given above took it granted that $Y=aX+b$ is of normal distribution if $X$ is of the same. Below we'll show that this the case.

If we have a function $y=ax+b$ then the inverse is

$$x=\phi(y)=\frac{y-b}{a}.$$

Now, following the OP's suggestion with this $\phi(y)$, we get

$$f_Y(y) = f_X\left(\frac{y-b}a\right)\frac1a=\frac 1{a\sqrt{2\pi(\sigma^2+\eta^2})}e^{\frac{\left(\frac{y-b}a-\mu\right)^2}{2\left(\sigma^2+\eta^2\right)}}.$$ We want

$$\frac 1{\sqrt{(\sigma^2+\eta^2})}\frac1a=\frac1{\sigma}.$$

Thus

$$ a=\frac{\sigma}{\sqrt{(\sigma^2+\eta^2})}$$

and, also

$$\frac{\left(\frac{y-b}a-\mu\right)^2}{2\left(\sigma^2+\eta^2\right)}=\frac{(y-b-a\mu)^2}{2a^2(\sigma^2+\eta^2)}=\frac{(y-\mu)^2}{2\sigma^2}$$

where $a$ works fine for the denominator and we need to secure that

$$b+a\mu=\mu.$$

for the numerator.

So,

$$b=\mu(1-a).$$