 Solution

ID: past-exam-of-the-options-ii-course-of-the-university-of-oxford/2012/m/m4/c/solution

ε > 0

. On the event

∣ Y_{n} - Y ∣ \leq ε

Y \leq x - ε

implies

Y_{n} \leq x

, while

Y_{n} \leq x

implies

Y \leq x + ε

. Hence

F_{Y} (x - ε) - P (∣ Y_{n} - Y ∣ > ε) \leq F_{Y_{n}} (x) \leq F_{Y} (x + ε) + P (∣ Y_{n} - Y ∣ > ε) .

Taking lower and upper limits in

n

removes the error probabilities. If

x

F_{Y}

, letting

ε ↓ 0

squeezes both bounds to

F_{Y} (x)

. Thus

Y_{n} \to Y in probability ⟹ Y_{n} \to Y in distribution .

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