Past exam of the mathematics course of the University of Oxford / Undated / Prelims / Mathematics III resit / 5

For the homogeneous linear recurrence relation, try $u_n=r^n$. Its characteristic equation of a linear recurrence isThe distinct roots of a polynomial $1,-4$ give the complementary terms $A+B(-4{)}^n$. A constant cannot absorb the forcing because $1$ is already a root of a polynomial of the characteristic equation; try a particular solution $cn$. Substitution gives $c=-4c+1$, hence $c=1/5$. The general solution is $u_n=A+B(-4{)}^n+n/5$.

The initial conditions give $A+B=0$ and $A-4B=4/5$. Hence $A=4/25$, $B=-4/25$, and

The homogeneous characteristic equation of a linear recurrence isThe repeated characteristic root of a linear recurrence $-1$ supplies the independent terms $(-1{)}^n$ and $n(-1{)}^n$. A constant particular solution is $1/2$, since $1/2=-2(1/2)-1/2+2$. Thus $u_n=1/2+(A+Bn)(-1{)}^n$.

The scan prints $u_2=0$, not $u_1=0$. These initial conditions are consistent. They give $A=1/2$ from $u_0=1$, then $0=1/2+A+2B$, so $B=-1/2$. ThereforeIn particular $u_1=1/2$ and $u_2=0$. Direct substitution verifies the recurrence relation. This is an example of nonconsecutive initial values for a linear recurrence; replacing the printed subscript by 1 would change the answer.

For random variables with finite second moments, their covariance isThe Cauchy-Schwarz inequality ensures the product is integrable. Expand and use linearity of expectation:The same identity holds whenever the quantities involved exist with sufficient integrability.

An expected value of a signed random variable is finite precisely when its absolute value is integrable. With the standard normal distribution probability density function,For $\alpha <1/2$, this integral equals $1/[\sqrt{2\pi }(1/2-\alpha )]<\infty$. For $\alpha \ge 1/2$, its integrand does not decay sufficiently and the integral diverges. ThereforeThe value is zero by the odd function and the symmetric distribution. For the other $\alpha$, both positive and negative parts have infinite integrals; formal cancellation would give only a Cauchy principal value, not an expected value. This distinction is the integrability threshold for exponential powers of a Gaussian.

Absolute integrability now requiresFor $\alpha \le 0$, the integrand is bounded by $z{e}^{-z^2/2}$, whose integral is finite. For $\alpha >0$, once $z^2\ge 1/\alpha$, the exponent is at least $\alpha z^4/2$, so the integral diverges. HenceAgain the zero follows from the odd function and symmetry; outside the integrable range both signed parts diverge, so the expected value is undefined.

Let $A=\{|Z|>2\}$ and $B=\{Z>0\}$. The standard normal distribution is symmetric and assigns no probability mass to zero. Thus $\mathbb{P}(B)=1/2$, and its equal two-sided tails giveThereforeso the two events are independent. The permitted approximation gives $\mathbb{P}(A)\approx 0.05$ and $\mathbb{P}(A\cap B)\approx 0.025$, but independence is exact.