 Solution

ID: past-exam-of-the-probability-statistics-and-options-ii-course-of-the-university-of-oxford/2011/m/m4/d/solution

For

t \geq 0

, the geometric survival probability gives

Pr (T_{k} > t) = Pr (S_{1} > k t) = (1 - \frac{1}{k})^{⌊ k t ⌋} ⟶ e^{- t} .

For

t < 0

, the distribution functions are zero, as is the limiting exponential distribution function. Since that limiting distribution function is continuous everywhere, these limits prove

T_{k} d Exp (1) .

The parameter is rate one, so the limiting mean is one, consistent with

E (S_{1} / k) = 1

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