Past exam of the year 1 of the computer science course of the University of Oxford / 2015 / Discrete mathematics, continuous mathematics, and probability (resit) / 9 / a

A probability density function must be nonnegative and have definite integral one. HereThereforeThis value is positive, so the nonnegativity requirement also holds.

The cosine function is positive on $(0,\pi /2)$ and negative on $(\pi /2,2)$. Thus any nonzero $\alpha$ makes $\alpha \cos x$ negative on an interval of positive length. Taking $\alpha =0$ makes the total definite integral zero. HenceNormalization alone would suggest $\alpha =1/\sin 2$, but that candidate still violates nonnegativity and therefore is not a probability density function.

We need $0<\alpha <2$. Indeed, if $\alpha <0$, the function is negative on $(\alpha ,0)$; if $\alpha =0$, the improper integral diverges at zero; and if $\alpha \ge 2$, its support interval is empty. For $0<\alpha <2$,Setting this equal to one gives the unique admissible valueOn this interval $1/x>0$, so both requirements for a probability density function are met.