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Mathematics
Set Theory, Logic, Probability, Statistics
How to change the support of a probability density function?
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[QUOTE="Stephen Tashi, post: 6349524, member: 186655"] did you mean ##\frac{(x-u)^{p-1} (v-x)^{q-1}}{Beta(p,q)(v-u)^{p+q-1}}## ? The phrase "make the transformation" is ambiguous even if you specify the transformation of coordinates has the form ##x' = Ax + B## and that ##[a,b]## is transformed to ##[u,v]##. Given a probability density ##f(x)## and a transformation of coordinates ##T(x) = Ax + B## such that ##T(a) = u ## and ##T(b) = v##, I think what you want is to define a probability density ##g(x)## [B]satisfying the requirement[/B] that ##\int_{u_1}^{v_1} g(x) dx = \int_{T^{-1}(u_1)}^{T^{-1}(v_1)} f(x) dx ## for all intervals ##[u_1,v_1]## The ##g(x)## satisfying that requirement is ##g(x) = (1/A) f(T^{-1}(x)) = (1/A) f(\frac{x-B}{A})## In the integral ##\int_{T^{-1}(u_1)}^{T^{-1}(v_1)}(1/A) f(\frac{x-B}{A}) dx ## make the change of variables ##y = T^{-1}(x) = \frac{x-B}{A} ## , ##dy = \frac{dx}{A}##, ##dx = A\ dy##. The integral is transformed to ##\int_a^b (1/A) f(y)\ A dy = \int_a^b f(y)dy = \int_a^b f(x) dx## (If ##T## was a more complicated transformation the choice ##g(x) = (1/A) f(T^{-1}(x)## might not work since a substitution might not produce the simple relation ##dx = A\ dy##) The conditions: ##T(a) = u = Aa + B## ##T(b) = v = Ab + B## imply ##A = \frac{v - u}{b-a}## ##B = u - Aa = u - \frac{v-u}{b-a}(a) ## [/QUOTE]
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Mathematics
Set Theory, Logic, Probability, Statistics
How to change the support of a probability density function?
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