Transformation rule

Theorem taught in Functions of random variables and limit theorems, part of Probability

Transformation rule

Type Theorem
Taught in Functions of random variables and limit theorems
Part Probability
Worked examples 4
Questions using it 1

Let \mathbf{X} take values in an open set of \mathbb{R}^n with density f_{\mathbf{X}}, and let \mathbf{Z}=g(\mathbf{X}) where g is a smooth bijection onto its image, with inverse g^{-1} and Jacobian determinant J that does not vanish. The density of \mathbf{Z} is

f_{\mathbf{Z}}(\mathbf{z})=f_{\mathbf{X}}\bigl(g^{-1}(\mathbf{z})\bigr)\,\bigl|\det Dg^{-1}(\mathbf{z})\bigr|.

Equivalently, if the Jacobian is computed for g rather than for the inverse, divide by |\det Dg| evaluated at the preimage. For a linear map \mathbf{Z}=A\mathbf{X} with A invertible, f_{\mathbf{Z}}(\mathbf{z})=f_{\mathbf{X}}(A^{-1}\mathbf{z})/|\det A|.

Examples

The square of a standard normal

Worked example from Functions of random variables and limit theorems

Let X\sim\operatorname{N}(0,1) and Y=X^2.

Question. Find the density of Y.

Solution. The map x\mapsto x^2 is not monotone, so work with the distribution function. For y>0,

F_Y(y)=\mathbb{P}(-\sqrt y\le X\le\sqrt y)=\Phi(\sqrt y)-\Phi(-\sqrt y)=2\Phi(\sqrt y)-1.

Differentiating, with \Phi'=\varphi the standard normal density,

f_Y(y)=2\varphi(\sqrt y)\cdot\frac1{2\sqrt y}=\frac{1}{\sqrt{2\pi}}\,y^{-1/2}e^{-y/2},\qquad y>0.

That is the \operatorname{Gamma}(1/2,1/2) density, the chi-squared law with one degree of freedom, since \Gamma(1/2)=\sqrt\pi.

Sum and proportion of two exponentials

Question page: Sum and proportion of two exponentials

Let X and Y be independent and exponential with rate 1. Set U=X+Y and V=X/(X+Y), so that X=UV and Y=U(1-V).

  1. Find the absolute value of the Jacobian determinant of (u,v)\mapsto(x,y).

  2. Find the joint density of (U,V).

  3. Show that U and V are independent, and name their marginal distributions.

  1. x=uv and y=u(1-v). The Jacobian matrix is \begin{pmatrix} v & u \\ 1-v & -u \end{pmatrix}, with determinant -u. Its absolute value is u for u>0.

  2. The joint density of (X,Y) is e^{-(x+y)}=e^{-u} on the positive quadrant. Multiplying by the absolute Jacobian gives f_{U,V}(u,v)=u e^{-u} for u>0 and 0<v<1.

  3. Integrating in u leaves f_V(v)=1 on [0,1]. Integrating in v leaves f_U(u)=u e^{-u} for u>0, which is \operatorname{Gamma}(2,1) in the shape–rate parameterisation. The joint density is the product, so U and V are independent. V is uniform on [0,1].

A quadratic function of a uniform variable

Question page: A quadratic function of a uniform variable

Let X be uniform on [1,3] and set Y=X^2-4. Find the density of Y, including the set on which it is zero.

On [1,3] the map x\mapsto x^2-4 increases from -3 to 5. For y in that range,

\mathbb{P}(Y\le y)=\mathbb{P}\bigl(X\le\sqrt{y+4}\bigr)=F_X(\sqrt{y+4}).

The uniform density on [1,3] is 1/2, so

f_Y(y)=\frac{1}{2}\cdot\frac{1}{2\sqrt{y+4}}=\frac{1}{4\sqrt{y+4}}, \qquad -3<y<5,

and f_Y(y)=0 otherwise.

Distance to the centre

Question page: Distance to the centre

A point is drawn uniformly from the disc of radius 4. Let R be its distance to the centre.

  1. Find \mathbb{P}(R>2).

  2. Find the density of R.

  1. The disc has area 16\pi and the disc of radius 2 has area 4\pi, so \mathbb{P}(R>2)=(16\pi-4\pi)/(16\pi)=3/4.

  2. For 0<r<4, \mathbb{P}(R\le r)=\pi r^2/(16\pi)=r^2/16. Differentiating gives f(r)=r/8 on (0,4), and f(r)=0 elsewhere.

Used in

See also

Multivariate normal distribution · Affine images of normal vectors · Normal linear model · Law of large numbers · Central limit theorem

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