Past exam of the year 1 of the computer science course of the University of Oxford / 2014 / Discrete mathematics, continuous mathematics and probability / 9

A probability density function must be nonnegative and integrate to one. For $c>0$,At $c=0$ the function is identically zero, while $c<0$ makes it negative on $x>1$. Thus $\boxed{c>0}$ is necessary and sufficient. The resulting law is a Pareto distribution with scale one and shape $c$.

Integrating the probability density function from the lower endpoint gives the cumulative distribution functionThe cumulative distribution function is continuous at $x=1$ and tends to one at infinity.

For $c>1$, the expected value is the convergent improper integralFor completeness, when $0<c\le 1$ the same positive integral diverges and the expected value is infinite.

The logarithm is strictly increasing, so $\mathbb{P}(Y\le y)=\mathbb{P}(X\le e^y)$. Since $X>1$ almost surely,Thus the logarithm of a Pareto random variable has an exponential distribution with rate $c$. Its probability density function value at the single endpoint can be chosen arbitrarily without changing the law.

By linearity of expectation and independence,For $\sigma >0$, apply Chebyshev's inequality at threshold $h\sigma$ and take the complementary event:If $\sigma =0$, every $X_i=\mu$ almost surely, so $W=\mu$ almost surely and the bound holds directly. This handles the zero-threshold case without applying Chebyshev's inequality at an inadmissible threshold.

Let $X_i$ be the head indicator random variable for toss $i$. These are independent and identically distributed random variables with a Bernoulli distribution of parameter $1/2$, so $\mu =1/2$, $\sigma =1/2$, and $W=N/n$. Taking $h=1/5$ makes $h\sigma =1/10$. The preceding Chebyshev's inequality bound givesThe endpoints are included because this event is exactly $|W-1/2|\le 1/10$.