Past exam of the options II course of the University of Oxford / 2012 / M / M4

Each ticket is equally likely, so the supplied sums giveHenceRepeated numerical values still represent different equally likely tickets; they must retain their multiplicities.

Replacement and mixing make the one hundred draws independent and identically distributed. Writing $S={\sum }_{j=1}^{100}{X}_j$, linearity of expectation and additivity of variances of independent variables give

The event $S\ge 500$ is contained in $|S-400|\ge 100$. Thus Chebyshev's inequality givesThe variance bound is rigorous and uses no normal approximation.

The central limit theorem suggests approximating $S$ by a normal variable with mean $400$ and variance $560$. ThusSince $S$ is integer-valued, the usual continuity correction replaces $500$ by $499.5$:These are approximation formulas rather than consequences of Chebyshev's bound; the standard deviation is $\sqrt{560}$, not $560$.

Convergence in probability to $Y$ means that for every $\epsilon >0$,Here the variables must be defined on a common probability space so that their differences are meaningful.

Convergence in distribution to $Y$ means convergence of cumulative distribution functions at every continuity point of the limiting distribution function:Convergence at a jump of $F_Y$ is not required. Only the laws are involved, so this definition does not require a particular coupling of the variables.

Assume convergence in probability and fix $\epsilon >0$. On the event $|Y_n-Y|\le \epsilon$, $Y\le x-\epsilon$ implies $Y_n\le x$, while $Y_n\le x$ implies $Y\le x+\epsilon$. HenceTaking lower and upper limits in $n$ removes the error probabilities. If $x$ is a continuity point of $F_Y$, letting $\epsilon \downarrow 0$ squeezes both bounds to $F_Y(x)$. Thus