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I guess the price of a Zero-Coupon Bond with infinite maturity should go to zero, what about its yield? I am asking this because I was dealing with the yield curve and its asymptotic properties when $t\to\infty$

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This is something that banks don't do very well (in my opinion), but we can look to the insurance industry for help.

  • Insurance liabilities often span decades, and the regulation has come up with something called the Ultimate Forward Rate (or UFR). It's currently a hotly debated topic with the advent of Solvency II (insurance regulation) coming into effect on 01/01/2016. This is because the UFR is not always set with an eye to long term interest rates, but more by looking at an appropriate liability discount rate.
  • The insurance industry preferred curve fitting approach is the Smith-Wilson model, which has the UFR as an input.

Hopefully this is a useful starting point for your research.

In the end, the actual value of the yield of an infinitely lived bond is irrelavant. As long as your infinite-year forward rate is reasonable (i.e. not $ \infty $), then $\lim_{t \rightarrow \infty} e^{-rt} = 0$ anyway.

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  • $\begingroup$ Thanks for the answer. Anyway, I’m not really interested in the price of such ZCB but in its yield (return) per-se. $\endgroup$ – fni Feb 26 '15 at 11:37
  • $\begingroup$ This is quite old, but a handy tool. CEIOPS (now called EIOPA, the European Insurance and Occupational Pensions Authority) says the UFR is 4.2%. ec.europa.eu/internal_market/insurance/docs/solvency/qis5/… $\endgroup$ – crunch Feb 26 '15 at 12:32
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while it is true that $$\lim_{T\to\infty} Z(t, T) = \lim_{T\to\infty} e^{-r(T-t)} = 0$$ this is when $r$ is independent of time to maturity, a flat and constant yield curve. In practice, we use yield curves which vary depending on what day they are estimated and what maturity the ZCB is. If in fact $r(t, T)$ depends on today and the maturity then the properties of that function are going to determine what the limit is. Of course, any model that allows for a non-zero price for an infinite maturity ZCB is admitting arbitrage.

Commonly, the Nelson Siegel and Nelson Siegel Svensson (original paper) models are used, in that case $$ r(t, T) = \beta_0 + \beta_1{1-\exp(-(T-t)/\tau)\over(T-t)/\tau} + \beta_2\left({1-\exp(-(T-t)/\tau)\over(T-t)/\tau}-\exp(-(T-t)/\tau)\right)$$

In the case of this model $\lim_{T\to\infty}r(t, T) = \beta_0$ whenever $\tau > 0$ and so $$\lim_{T\to\infty}Z(t, T) = \lim_{T\to\infty} e^{-r(t, T)(T-t)}=0$$ whenever $\beta_0 > 0$

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