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Show that P(x|π) = π^x(1-π)^(1-x) can be written in exponential form.

User Ignar
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Final answer:

P(x|\pi) is expressed in exponential form by default for x = 0 or x = 1; if seeking a general exponential expression, the function \pi^x can be rewritten using the natural log and exponential function as e^{x ln(\pi)}. The distribution function uses exponential form with m being the decay rate (m = 1/\mu).

Step-by-step explanation:

To show that the function P(x|\pi) = \pi^x(1-\pi)^{1-x} can be written in exponential form, observe that the function is already in that form if we substitute x with the binary outcomes 0 or 1. When x equals 1, the function simplifies to P(1|\pi) = \pi. Conversely, when x equals 0, it simplifies to P(0|\pi) = 1 - \pi. However, if the goal is to express \pi^x for any arbitrary x as an exponential function of x, we could resort to the natural logarithm and the exponential function's inverse nature to rewrite \pi^x as e^{x \ln(\pi)}, which would then express \pi^x in a purely exponential format using the base e of the natural logarithm.

Regarding the cumulative distribution function for the exponential distribution, P(X \leq x) = 1 - e^{-mx} where m is the decay rate, for a continuous random variable X which follows an exponential distribution with mean \mu, we note that m is related to the mean by m=1/\mu.

User Crystark
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