For a distribution function F(x), define F(y) = inf{t: F(t) ≥ y} F (y)inf{t: F(t) > y}. We know F (y) is left-continuous. Show F, (y) is right continuous and show λ{ue (0, 1]: F (u) #F, (u)} = 0, where, as usual, λ is Lebesgue measure. Does it matter which inverse we use?

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For a distribution function F(x), define
F(y) = inf{t: F(t) ≥ y}
F(y)inf{t: F(t) > y}.
We know F (y) is left-continuous. Show F, (y) is right continuous and
show
λ{u € (0, 1]: F (u) #F, (u)} = 0,
where, as usual, λ is Lebesgue measure. Does it matter which inverse we
use?
Transcribed Image Text:For a distribution function F(x), define F(y) = inf{t: F(t) ≥ y} F(y)inf{t: F(t) > y}. We know F (y) is left-continuous. Show F, (y) is right continuous and show λ{u € (0, 1]: F (u) #F, (u)} = 0, where, as usual, λ is Lebesgue measure. Does it matter which inverse we use?
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