I have been struggling with the following question:
Consider the yield to maturity of a static portfolio of coupon bonds, each with a finite expiry date, in which shorted bonds are allowed (so the total of the coupon payments as well as the possible face value payments, at each time can be negative, positive or zero). Suppose the first payments on this portfolio will take place at time at time $T_{1} > 0$, where time $t = 0$ denotes the present time and suppose that the value of the portfolio is positive at the present time and less than the total sum of the (undiscounted) cash-flow payments that the portfolio will generate.
(a) Show that the yield to maturity y is uniquely defined in case the duration $D = D(y)$ is positive for all $y > 0$.
(b) Give an example of such a static portfolio with (at least) two yield-to maturity values.
So far I have come up with the following:
Yield to maturity is $y$ that solves the following equation:
$ p(t) = \sum_{i = 1}^{n} c_{i} e^{-y T_{i}} $
where $p(t)$ is the market value of the portfolio, $c_{i}$ are net coupon payments and time i, and I have incorporated face value payments into the $c_{i}s$ for simplicity. The question specifies that the following must hold:
$0 < p(t) < \sum_{i = 1}^{n} c_{i} $.
The duration is given by:
$ D = \frac{\sum_{i = 1}^{n} T_{i} c_{i} e^{-y T_{i}}}{p} $
and this must be positive for every y > 0.
I know that $\sum_{i = 1}^{n} T_{i} c_{i} e^{-y T_{i}} > 0 $ must hold for this to be true, but am at a loss as to how to proceed after this.
If anyone could assist me or even point me in the right direction I would be very grateful.
For the first part: Consider the function $$f(y)=\sum {c_ie^{-yT_i}}$$ We have that $f(0)>p$, and since $f'(y)<0$ for all $y>0$ there is at most one $y$ such that $f(y)=p$. Since $f(y)\to 0$ as $y\to\infty$ it follows that at some $y$ we have $f(y)<p$, so by the intermediate value theorem there is a unique $y$ with $f(y)=p$. This is the unique yield to maturity.
For the second part, we need $f(y)$ to cross $p$ twice. The polynomial $x(x-\frac12)(x-\frac14)$ is $x^3-\frac34 x^2+\frac18x$. We could choose $T_1=1$, $T_2=2$, $T_3=3$, $c_1=0.125$, $c_2=-0.75$, and $c_3=1$. Then $y=-\ln 0.25$ and $y=-\ln 0.5$ are both yields to maturity if the price is zero.