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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
FY(y)=P(g(X)≤y)F_Y(y) = P(g(X)\le y)

then differentiate

Monotone transform
fY(y)=fX(g−1(y))∣ddyg−1(y)∣f_Y(y) = f_X\big(g^{-1}(y)\big)\left|\dfrac{d}{dy}g^{-1}(y)\right|
Maximum of n iid
Fmax⁡(x)=[F(x)]nF_{\max}(x) = \left[F(x)\right]^{n}
Minimum of n iid
Smin⁡(x)=[S(x)]nS_{\min}(x) = \left[S(x)\right]^{n}
Convolution
fX+Y(s)=∫fX(x)fY(s−x) dxf_{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)≤y)P(g(X)\le y) can always be rewritten as a probability about XX when gg 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 xx exactly when ALL of them are, so the CDFs multiply; the minimum is above xx exactly when all of them are, so the SURVIVAL functions multiply.

Worked example

Let $U$ be uniform on (0,1) and $Y = -2\ln U$. Find $P(Y > 3)$.

  1. Y>3  ⟺  −2ln⁡U>3  ⟺  ln⁡U<−1.5  ⟺  U<e−1.5Y > 3 \iff -2\ln U > 3 \iff \ln U < -1.5 \iff U < e^{-1.5}.
  2. The inequality flipped when dividing by −2-2 - the step most often missed.
  3. P(U<e−1.5)=e−1.5P(U < e^{-1.5}) = e^{-1.5} because UU is standard uniform.
  4. P(Y>3)=e−1.5=0.2231P(Y>3) = e^{-1.5} = 0.2231. In fact YY is exponential with mean 2.

Answer: 0.2231

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

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