Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 43 / 5

For $0\le s\le 1$, pointwise concavity of the utility function givesTaking expectations is legitimate because all three values are finite. Therefore $\boxed{F(s\theta +(1-s)\eta )\ge sF(\theta )+(1-s)F(\eta )}$: $F$ is concave. Strict concavity is not required and is not asserted.

For secant domination for expected utility derivatives, at each sample value $x$ set $g_x(\theta )=U(\theta x)$. This is a differentiable concave function, regardless of the sign of $x$. Fix ${\theta }_0$ and $\delta >0$. For $\theta \in [{\theta }_0-\delta ,{\theta }_0+\delta ]$, concavity bounds its derivative between the two outer secant slopes:Because $U<0$, finiteness of $F$ means $\mathbb{E}|U(\eta X)|<\infty$ at each of these four endpoints. Their absolute values divided by $\delta$ therefore give a common integrable bound on $|g_X^{\prime }(\theta )|$. Difference quotients satisfy the same bound by the mean value theorem. The dominated convergence theorem permits differentiation under the expectation, and pointwise continuity of $X{U}^{\prime }(\theta X)$ with the same bound gives continuity of the derivative. Hence $\boxed{F^{\prime }(\theta )=\mathbb{E}[X{U}^{\prime }(\theta X)]}$ and $F\in C^1(\mathbb{R})$. This argument proves absolute integrability of the displayed derivative; no unjustified differentiation assumption is needed.

A strictly increasing differentiable concave utility function on all of $\mathbb{R}$ has $U^{\prime }(x)>0$ everywhere. Otherwise monotonicity and the decreasing derivative would make it constant on a right half-line. Its tangent at zero also shows $U(x)\to -\infty$ as $x\to -\infty$.

If $X$ has only one sign, the stated second alternative already holds. Otherwise both $\mathbb{P}(X<0)$ and $\mathbb{P}(X>0)$ are positive. As $\theta \to \infty$, $-U(\theta X)$ tends to infinity on $\{X<0\}$; as $\theta \to -\infty$, it does so on $\{X>0\}$. Since $-U\ge 0$, Fatou lemma implies $F(\theta )\to -\infty$ at both ends. Continuity gives a finite maximizer ${\theta }_{\ast }$, and differentiability gives $F^{\prime }({\theta }_{\ast })=0$.

For optimal marginal utility as a one-period pricing density, put $Z=U^{\prime }({\theta }_{\ast }X)$. It is strictly positive. On $\{|X|\ge 1\}$ it is bounded by $|X|U^{\prime }({\theta }_{\ast }X)$, whose expectation is finite by part ii. On $\{|X|<1\}$ continuity bounds $U^{\prime }$ on the compact interval $[-|{\theta }_{\ast }|,|{\theta }_{\ast }|]$. Thus $Z$ is integrable and $XZ$ is absolutely integrable. Consequently $\boxed{\mathbb{E}[XZ]=0}$. Normalizing $Z$ by its positive expectation, if desired, makes it a probability density with the same zero pricing expectation.

Under a proportional transaction cost, the sample gain $h_x(\theta )=\theta x-\epsilon |\theta |$ is concave when $\epsilon \ge 0$. The utility function is increasing and concave, soTake finite expectations. Thus $\boxed{G\text{ is concave}}$. The monotonicity of $U$ is essential to composing it with the concave transaction-cost payoff.

On each open half-line the sample objective is a utility of a linear gain: $U(\theta (X-\epsilon ))$ on the positive half-line, and $U(\theta (X+\epsilon ))$ on the negative half-line. On a compact subinterval wholly inside either half-line, the secant domination argument from the frictionless part ii applies verbatim, using finiteness of $G$ at four surrounding points. The dominated convergence theorem therefore gives the derivativesThey are continuous on their respective half-lines. Near zero, the same outer secants of the globally concave sample objective bound its one-sided slopes. Dominated convergence consequently givesHere $\mathbb{E}|X|<\infty$: the frictionless derivative at zero already proves it, since $U^{\prime }(0)>0$. Thus zero generally has a transaction-cost kink; when $\epsilon =0$ the two formulas agree.

For marginal utility pricing with proportional transaction costs, if $X-\epsilon \ge 0$ almost surely or $-X-\epsilon \ge 0$ almost surely, the source's second alternative holds. Otherwise $\mathbb{P}(X<\epsilon )>0$ and $\mathbb{P}(X>-\epsilon )>0$. The Fatou lemma argument now gives $G(\theta )\to -\infty$ at both ends, since a large positive holding loses on the first event and a large negative holding loses on the second. Let ${\theta }_{\ast }$ be a finite maximizer.

If ${\theta }_{\ast }>0$, set $Z_0=U^{\prime }({\theta }_{\ast }(X-\epsilon ))$. The derivative condition gives $\mathbb{E}[X{Z}_0]=\epsilon \mathbb{E}{Z}_0$. As before, split according to $|X-\epsilon |\ge 1$ to obtain integrability of $Z_0$ from the absolutely integrable derivative, with a compact-set bound on the complement. If ${\theta }_{\ast }<0$, use $Z_0=U^{\prime }({\theta }_{\ast }(X+\epsilon ))$ and obtain $\mathbb{E}[X{Z}_0]=-\epsilon \mathbb{E}{Z}_0$. In either case normalize $Z=Z_0/\mathbb{E}{Z}_0$; then $Z>0$, $\mathbb{E}Z=1$, and $XZ$ is absolutely integrable.

If ${\theta }_{\ast }=0$, a concave maximum has $G_{-}^{\prime }(0)\ge 0\ge G_{+}^{\prime }(0)$, so $\mathbb{E}X\in [-\epsilon ,\epsilon ]$. Choose $Z=1$. In every case of this first alternative, $\boxed{\mathbb{E}[XZ]\in [-\epsilon ,\epsilon ]}$. The normalization establishes a stronger, economically meaningful statement than the literal unnormalized condition: the expected gain under the resulting density lies inside the transaction-cost spread.