Talk:Truncated distribution

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I would like to add the argument that since f(x | t) is a truncation of f(x) such that x < t, we know that:

f(x|t)= \frac{f(x)}{F(t)}.

By Bayes Rule:

g(t|x)= \frac{f(x|t)g(t)}{f(x)}

which reduces to:

g(t|x)= \frac{g(t)}{F(t)} = \frac{g(t)}{\int_{-\infty}^t[\int_{x}^{\infty} f(x|t)g(t)dt]dx}

However, this does not work. t does not truncate the distribution of x so much as define the distribution. In light of this, this page probably should be somewhere else.