- #1
Juanriq
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Ahoy hoy, trying to see if I am on the right track with a question.
Is this right so far? Thanks!
Homework Statement
Let [itex] T: (X, \mathcal{B}, m)[/itex] to itself be an mpt of a probability space. Suppose [itex] X = X_1 \cup X_2 [/itex] where [itex] X_1 \cap X_1 = \emptyset [/itex] and [itex] m(X_i) > 0 [/itex] and also [itex] T^{-1}X_i = X_i [/itex]. Let [itex] T: (X_i, \mathcal{B}|X_i, m_i) [/itex] to itself where [itex] T_i = T|X_i [/itex] and [latex[ m_i = m|X_i [/itex] normalized. What is the relation between [itex] h_m(T) [/itex] and [itex] h_{m_i}}(T_i)[/itex]?Homework Equations
The Attempt at a Solution
I know [itex] m(X) = 1 [/itex], so say [itex] m(X_1) = p [/itex] and [itex] m(X_2) = 1-p [/itex]. This would mean that [itex] m_1 = \frac{m|X_1}{m(X_1)} = \frac{m|X_1}{p} [/itex] and also [itex] m_2 = \frac{m|X_2}{m(X_2)} = \frac{m|X_1}{1-p} [/itex]. Now, [itex]h_m(T) = -m(X_1) \log m(X_2) - m(X_2) \log m(X_2) [/itex] and so [itex] h_{m_i}}(T_i) = -m_1(X_1) \log m_1(X_2) - m_2(X_2) \log m_2(X_2)[/itex].Is this right so far? Thanks!