Exam PApplicationsFree to read

Transformations and order statistics

Push a random variable through a function: use the CDF method when you can, the Jacobian when you must, and treat maxima and minima as CDF products.

The formulas

CDF method

$$F_Y(y)=P(g(X)\le y)$$

then differentiate

Monotone transform

$$f_Y(y)=f_X(g^{-1}(y))\left|\frac{d}{dy}{g}^{-1}(y)\right|$$

Maximum of n iid

$$F_{\max }(x)={\left[F(x)\right]}^n$$

Minimum of n iid

$$S_{\min }(x)={\left[S(x)\right]}^n$$

Convolution

$$f_{X+Y}(s)=\int f_X(x)f_Y(s-x)dx$$

Where it comes from

1. The CDF method works because $P(g(X)\le y)$ can always be rewritten as a probability about $X$ when $g$ is invertible on the relevant region.
2. Differentiating that identity produces the Jacobian factor — the absolute derivative of the inverse.
3. The maximum is below $x$ exactly when ALL of them are, so the CDFs multiply; the minimum is above $x$ exactly when all of them are, so the SURVIVAL functions multiply.

Worked example

This answer is recomputed from the site’s own interest-theory and probability functions every time the test suite runs, so the page and the mathematics cannot drift apart.

Memory hooks

• Always go through the CDF. The density method is a shortcut that only works for monotone g.
• Max ⇒ multiply CDFs. Min ⇒ multiply survival functions. Nothing else to remember.

Traps

• Forgetting the absolute value on the Jacobian for a decreasing transform.
• Failing to flip the inequality when dividing by a negative number.

Related

• Random variables, CDFs and survival functions
• Exponential distribution
• Continuous uniform distribution

Drill this: the Exam P question bank has original questions on this topic, and today’s free round is open to everyone.