# LOTUS

by Daniel Pollithy

If you want to calculate the expected value of a continuous random variable which was transformed by a monotonic function, then the law of the unconscious statistician provides a convenient shortcut.

## Transforming a random variable

We have a continuous random variable $X$ with a probability density function $f_{X}(x)$. This could for example be our last knowledge about the position of a robot.

Now, we apply a continuous, monotonic function $g(\cdot)$ to the random variable. This could be a simple motion model for the robot.

If we want to calculate $E[g(X)]$ then LOTUS tells us that we don’t have to solve $g(X)$ but instead we can write:

$E[g(X)] = \int_{+\infty}^{+\infty}{g(x) \cdot f_{X}(x) dx}$

## Proof

For the proof let’s assume $g$ to be strictly increasing (decreasing would also be possible). And due to its continuity, $g(x)$ therefore has a positive derivative for every x. This results in $g(\cdot)$ being bijective therefore g is invertible. We call its inverse $g^{-1}$. With $g^{-1}: Y \rightarrow X$

### Prepare change of variable

The derivative of a function and its inverse are related. They are reciprocal:

$\frac{dx}{dy} \cdot \frac{dy}{dx} = 1$ $\frac{dx}{dy} = \frac{1}{\frac{dy}{dx}}$

First we replace $y$ with $g(x)$ on the right side:

$\frac{dx}{dy}= \frac{1}{\frac{d g(x)}{dx}}$

Second we replace $x$ with $g^{-1}(y)$ on the right side:

$\frac{dx}{dy}= \frac{1}{\frac{d g(g^{-1}(y))}{dx}}$

Multiplying with dy on both sides:

$dx = \frac{1}{\frac{d g(g^{-1}(y))}{dx}} dy$

### Expected value with exchanged variable

$\int_{+\infty}^{+\infty}{g(x) \cdot f_{X}(x) dx}$

And now we can switch from x to y. First, replace g(x) with y. Second, replace $f_{X}(x)$ with $f_{X}(g^{-1}(y))$. And third, replace dx with the right hand side from “dx = …” above:

$\int_{+\infty}^{+\infty}{g(x) \cdot f_{X}(x) dx} = \int_{+\infty}^{+\infty}{y \cdot f_{X}(g^{-1}(y)) \frac{1}{\frac{d g(g^{-1}(y))}{dx}} dy}$

We have now switched to integrating over y.

### Cumulative density function

$F_{Y}(y) = Pr(Y \le y)$

Apply g:

$F_{Y}(y) = Pr(g(X) \le y)$

Apply $g^{-1}$ on both sides

$F_{Y}(y) = Pr(X \le g^{-1}(y))$ $F_{Y}(y) = F_{X}(g^{-1}(y))$

### Derivative of CDF

Now we can get the derivative of $F_{Y}(y)$ for y in order to get $f_{Y}(y)$. The chain rule is used. Note that this is the place where we need the derivative of $g^{-1}(y)$ which is $\frac{1}{\frac{d g(g^{-1}(y))}{dx}}$ !

Apply the CDF solution from above:

$f_{Y}(y) = \frac{d}{dy} F_{Y}(y) = \frac{d}{dy} F_{X}(g^{-1}(y))$

Apply the chain rule of derivation:

$= f_{x}(g^{-1}(y)) \cdot \frac{d}{dy} g^{-1}(y)$ $= f_{X}(g^{-1}(y)) \cdot \frac{1}{\frac{d g(g^{-1}(y))}{dx}} =$

### Plug-in

Two sections before, we got to this point:

$E[g(X)] = \int_{+\infty}^{+\infty}{y \cdot f_{Y}(y) dy} = \int_{+\infty}^{+\infty}{y \cdot f_{X}(g^{-1}(y)) \frac{1}{\frac{d g(g^{-1}(y))}{dx}} dy}$

Looking at the last formula from the section “Prepare change of variable”, we find that $f_{X}(g^{-1}(y)) \cdot \frac{1}{\frac{d g(g^{-1}(y))}{dx}} dy$ to be the same as $f_{X}(g^{-1}(y)) dx$

$E[g(X)] = \int_{+\infty}^{+\infty}{y \cdot f_{X}(g^{-1}(y)) dx}$

Per definition $g^{-1}(y)$ can be replaced by x.

$E[g(X)] = \int_{+\infty}^{+\infty}{y \cdot f_{X}(x) dx}$

And also $y$ can be replaced by g(x).

$E[g(X)] = \int_{+\infty}^{+\infty}{g(x) \cdot f_{X}(x) dx}$