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suppose u is a uniform(0, 1) random variable. consider f-1(u), where f-1(.) is the inverse of the cdf of the random variable x. so, f-1(u) is a transformation of u. assume f-1(.) is strictly increasing. what is the distribution of f-1(u)?

Sagot :

The distribution of the given function is a uniform distribution.

A strictly increasing function is the one which increases continuously in a given interval. If f-1(u) is a strictly increasing function of u, then the distribution of f-1(u) is the same as that of u. This is because a strictly increasing function preserves the rank ordering of its input, so if u-1 and u-2 are two random variables with a uniform distribution on the interval (0,1), then the rank order of f-1(u-1) and f-1(u-2) will be the same as the rank order of u-1 and u-2. Since the uniform distribution is defined by the rank order of its values, this means that the distribution of f-1(u) is also uniform on the interval (0,1).

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