Editing Probability density function (section)

==Products and quotients of independent random variables==
{{See also|Product distribution|Ratio distribution}}

Given two independent random variables {{math|''U''}} and {{math|''V''}}, each of which has a probability density function, the density of the product {{math|1=''Y'' = ''UV''}} and quotient {{math|1=''Y'' = ''U''/''V''}} can be computed by a change of variables.

===Example: Quotient distribution===
To compute the quotient {{math|1=''Y'' = ''U''/''V''}} of two independent random variables {{math|''U''}} and {{math|''V''}}, define the following transformation:
<math display="block">\begin{align}
Y &= U/V \\[1ex]
Z &= V
\end{align}</math>

Then, the joint density {{math|''p''(''y'',''z'')}} can be computed by a change of variables from ''U'',''V'' to ''Y'',''Z'', and {{math|''Y''}} can be derived by [[marginalizing out]] {{math|''Z''}} from the joint density.

The inverse transformation is
<math display="block">\begin{align}
U &= YZ \\
V &= Z
\end{align}</math>

The absolute value of the [[Jacobian matrix]] determinant <math>J(U,V\mid Y,Z)</math> of this transformation is:
<math display="block">
\left| \det\begin{bmatrix}
\frac{\partial u}{\partial y} & \frac{\partial u}{\partial z} \\
\frac{\partial v}{\partial y} & \frac{\partial v}{\partial z}
\end{bmatrix} \right|
=
\left| \det\begin{bmatrix}
z & y \\
0 & 1
\end{bmatrix} \right|
= |z| .
</math>

Thus:
<math display="block">p(y,z) = p(u,v)\,J(u,v\mid y,z) = p(u)\,p(v)\,J(u,v\mid y,z) = p_U(yz)\,p_V(z)\, |z| .</math>

And the distribution of {{math|''Y''}} can be computed by [[marginalizing out]] {{math|''Z''}}:
<math display="block">p(y) = \int_{-\infty}^\infty p_U(yz)\,p_V(z)\, |z| \, dz</math>

This method crucially requires that the transformation from ''U'',''V'' to ''Y'',''Z'' be [[bijective]].  The above transformation meets this because {{math|''Z''}} can be mapped directly back to {{math|''V''}}, and for a given {{math|''V''}} the quotient {{math|''U''/''V''}} is [[monotonic]].  This is similarly the case for the sum {{math|''U'' + ''V''}}, difference {{math|''U'' − ''V''}} and product {{math|''UV''}}.

Exactly the same method can be used to compute the distribution of other functions of multiple independent random variables.

===Example: Quotient of two standard normals===
Given two [[standard normal distribution|standard normal]] variables {{math|''U''}} and {{math|''V''}}, the quotient can be computed as follows.  First, the variables have the following density functions:
<math display="block">\begin{align}
p(u) &= \frac{1}{\sqrt{2\pi}} e^{-{u^2}/{2}} \\[1ex]
p(v) &= \frac{1}{\sqrt{2\pi}} e^{-{v^2}/{2}}
\end{align}</math>

We transform as described above:
<math display="block">\begin{align}
Y &= U/V \\[1ex]
Z &= V
\end{align}</math>

This leads to:
<math display="block">\begin{align}
p(y) &= \int_{-\infty}^\infty p_U(yz)\,p_V(z)\, |z| \, dz \\[5pt]
&= \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2} y^2 z^2} \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2} z^2} |z| \, dz \\[5pt]
&= \int_{-\infty}^\infty \frac{1}{2\pi} e^{-\frac{1}{2}\left(y^2+1\right)z^2} |z| \, dz \\[5pt]
&= 2\int_0^\infty \frac{1}{2\pi} e^{-\frac{1}{2}\left(y^2+1\right)z^2} z \, dz \\[5pt]
&= \int_0^\infty \frac{1}{\pi} e^{-\left(y^2+1\right)u} \, du && u=\tfrac{1}{2}z^2\\[5pt]
&= \left. -\frac{1}{\pi \left(y^2+1\right)} e^{-\left(y^2+1\right)u}\right|_{u=0}^\infty \\[5pt]
&= \frac{1}{\pi \left(y^2+1\right)}
\end{align}</math>

This is the density of a standard [[Cauchy distribution]].