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02 Inferences about distributions of function(RV)

Y = g(X).

Use \(Pr(g(X) \in A) = Pr(X \in g^{-1}(A))\). So, given CDF, PDF of X, can deduce CDF of g(X) and thence derive PDF of g(X).

Using \(\frac{dg^{-1}(Y)\)

If g is monotone in \((x, x+\gd x)\): \(p_{X}(x)\gd x \approx p_{Y}(y)\gd y\), taking \((x, x+\gd x)\) to \((y, y+\gd y)\) using g: So \(p_Y(y) = p_{X}(x)|\frac{dx}{dy}| = p_{X}(g^{-1}(y))|\frac{d g^{-1}(y)}{dy}| \): so maximum probability density changes with variable change.

If g is not continuous, but \(\exists\) partition \(A_{0}, .. A_{k}\) with \(Pr(X \in A_{0}) = 0\), with \(\set{g_{i}} = g \) over \(\set{A_{i}}\) monotone; then \(p_Y(y) = \sum_{i} p_{X}(g^{-1}(y))|\frac{d g_{i}^{-1}(y)}{dy}|\); where \(\sum\) appears to account for the probability that Y=y over various domains of X.

Extension to multidimensional distributions

$$Y = g(X_{1}, X_{2}); \ X_{1} = h(Y, X_{2})\(. Fix \)X_{2} = x_{2}\(; get \)p(Y, x_{2}) = p_{X_{1}, X_{2}}(X_{1} \ = h^{-1}(Y, x_{2})| x_{2})|\frac{dh^{-1}(Y, x_{2})}{d Y}|\(; then do \)p_{Y}(y) = \int p(Y, x_{2}) dx_{2}$$.

Using moment generating functions

Given \(m_{X}(t)\), find \ \(m_{Y}(t) = E[e^{f(X)t}]\); thence deduce pdf of Y.