Editing Integration by substitution (section)

==Application in probability==

Substitution can be used to answer the following important question in probability: given a random variable {{mvar|X}} with probability density {{math|''p''<sub>''X''</sub>}} and another random variable {{mvar|Y}} such that {{mvar|1= ''Y''= ''ϕ''(''X'')}} for [[Injective function|injective]] (one-to-one) {{mvar|''ϕ'',}} what is the probability density for {{mvar|Y}}?

It is easiest to answer this question by first answering a slightly different question: what is the probability that {{mvar|Y}} takes a value in some particular subset {{mvar|S}}? Denote this probability {{math|''P''(''Y'' &isin; ''S'').}}  Of course, if {{mvar|Y}} has probability density {{math|''p''<sub>''Y''</sub>}}, then the answer is:
<math display="block">P(Y \in S) = \int_S p_Y(y)\,dy,</math>
but this is not really useful because we do not know {{math|''p''<sub>''Y''</sub>;}} it is what we are trying to find. We can make progress by considering the problem in the variable {{mvar|X}}. {{mvar|Y}} takes a value in {{mvar|S}} whenever {{mvar|X}} takes a value in <math display="inline">\phi^{-1}(S),</math> so:
<math display="block">P(Y \in S) = P(X \in \phi^{-1}(S)) = \int_{\phi^{-1}(S)} p_X(x)\,dx.</math>

Changing from variable {{mvar|x}} to {{mvar|y}} gives:
<math display="block">P(Y \in S) = \int_{\phi^{-1}(S)} p_X(x)\,dx = \int_S p_X(\phi^{-1}(y)) \left|\frac{d\phi^{-1}}{dy}\right|\,dy.</math>
Combining this with our first equation gives:
<math display="block">\int_S p_Y(y)\,dy = \int_S p_X(\phi^{-1}(y)) \left|\frac{d\phi^{-1}}{dy}\right|\,dy,</math>
so:
<math display="block">p_Y(y) = p_X(\phi^{-1}(y)) \left|\frac{d\phi^{-1}}{dy}\right|.</math>

In the case where {{mvar|X}} and {{mvar|Y}} depend on several uncorrelated variables (i.e., <math display="inline">p_X=p_X(x_1, \ldots, x_n)</math> and <math>y=\phi(x)</math>), <math>p_Y</math>can be found by substitution in several variables discussed above. The result is:
<math display="block">p_Y(y) = p_X(\phi^{-1}(y)) \left|\det D\phi ^{-1}(y) \right|.</math>