Probability
4. Advanced Distributions
Transformations

Transformations of Random Variables

Often you know the distribution of XX, but you care about a transformed variable:

Y=g(X).Y=g(X).

There are two standard tools:

  1. CDF method (works for discrete or continuous; especially good for monotone transforms)
  2. Change of variables / Jacobian (continuous PDFs)

CDF method (general)

For any yy:

FY(y)=P(Yy)=P(g(X)y).F_Y(y)=P(Y\le y)=P(g(X)\le y).

If you can rewrite the event g(X)yg(X)\le y in terms of XX, you can compute FY(y)F_Y(y), then differentiate (continuous case) to get fY(y)f_Y(y).

Continuous change-of-variables (monotone)

If Y=g(X)Y=g(X) is strictly monotone and differentiable, and XX has density fXf_X:

fY(y)=fX(g1(y))ddyg1(y). f_Y(y)= f_X(g^{-1}(y))\left|\frac{d}{dy}g^{-1}(y)\right|.

Linear transforms

If Y=aX+bY=aX+b:

  • E[Y]=aE[X]+bE[Y]=aE[X]+b
  • Var(Y)=a2Var(X)\mathrm{Var}(Y)=a^2\mathrm{Var}(X)
  • If XX is Normal, then YY is also Normal.

Test Your Knowledge

Example: Linear transformation of a Normal

Let XX be the temperature in Celsius, normally distributed with μ=20\mu = 20 and σ=5\sigma = 5. We apply a transformation to convert to Fahrenheit: Y=1.8X+32Y = 1.8X + 32. What is the distribution of YY?

View Step-by-Step Solution

A linear transformation of a Normal variable results in another Normal variable. We transform the mean and variance.

Mean: E[Y]=1.8(E[X])+32=1.8(20)+32=68E[Y] = 1.8(E[X]) + 32 = 1.8(20) + 32 = 68

Variance: Var(Y)=(1.8)2Var(X)=3.24×25=81\mathrm{Var}(Y) = (1.8)^2\,\mathrm{Var}(X) = 3.24 \times 25 = 81

So the standard deviation is 81=9\sqrt{81}=9 (which is 1.8×51.8\times 5).

Therefore, YN(68,81)Y \sim N(68, 81).