Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 208 / 1

Because the moment-generating function is finite in a neighborhood to the right of zero and $\mathbb{E}X=0$, its cumulant-generating function satisfiesDivide the defining Sub-Gamma random variable in the right tail inequality by ${\lambda }^2/2$ and let $\lambda \downarrow 0$. The right side converges to $\nu$, proving $\mathrm{Var}(X)\le \nu$.

Independence makes cumulant-generating functions additive:SetFor $0<\lambda <1/c$, every denominator is positive and $1-c_i\lambda \ge 1-c\lambda$, soAlso $\mathbb{E}Z=0$, and hence $Z\in {\Gamma }_{+}(\nu ,c)$.

The Chernoff bound gives, for $0<\lambda <1/c$,Writing $u=ct/\nu$, elementary differentiation shows that the exponent is minimized atSubstitution givesThus $h$ is the Legendre transform generated by the sub-Gamma cumulant bound after its natural rescaling.

Put $x=\sqrt{2\nu t}+ct$ and $y=c^{2}t/\nu$. ThenConsequentlyThe tail bound from part c therefore has exponent $-(\nu /c^2)y=-t$, and