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4

Set $X_t=B_t$ and $Y_t=B_t-B_s$, $Y_t\sim N(0,t-s)$ and $X_t$ and $Y_t$ are independent. $$I=P(B_t>0, B_s<0)=P(X_t>0\,, Y_t>X_t)$$ $$I=\frac{1}{2\pi\sqrt{t(t-s)}}\int_{0}^{\infty}\int_{x}^{+\infty}\exp\left(-\frac{x^2}{2t}-\frac{y^2}{2(t-s)}\right)dydx$$ Set $$x={\sqrt{t}}\,\,r\cos \theta$$ $$\quad y={\sqrt{t-s}}\,\,r\sin \theta$$ we have $$dy\...


2

There is no need to resort to negative-coupon bonds. A negative $R_t$ is simply a negative payment. For a simple example, build a portfolio consisting of being long a $n$ maturity bond paying a coupon C on $t \in \left\lbrace 1, 2, ..., n \right\rbrace$ and short a zero-coupon bond with face value $V > C$ maturing at $t^*$. Then, $R_t > 0$ for $t \ne ...


6

The answer is NO, with very few exceptions There might be bonds with negative coupon(s), and the Bloomberg search even finds some, but there are plenty of reasons why negative coupons are impractical. Instead of having negative coupons on the issue, there are bonds with low or 0 coupons, issued at a premium and having a negative yield. Here are some of ...


5

IMHO the problem isn't stated correctly indeed, in the sense that the Radon-Nikodym derivative provided as the "solution" is not the unique way to define a measure $\mathbb{Q}$ equivalent to $\mathbb{P}$ and under which $X_t$ is a martingale. Just take $$\frac {d\mathbb{Q}}{d\mathbb{P}} =\mathcal{E}\left(-\int_0^t \cos(s) dW_s + a\right)$$ for any $a \in \...



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