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I'm familiar with the expression for a (continuously compounded) credit spread of the form $$ c(t,T) = -\frac{1}{T-t} \ln \frac{v(t,T)}{p(t,T)},$$ where $p(t,T)$ denotes the time $t$ price of a $T$-maturity default-free bond, and $v(t,T)$ denotes the time $t$ price of a $T$-maturity defaultable bond. Using some standard assumptions from credit risk models (that the default time $\tau$ is independent of the short rate $r_t$ process so the $v(t,T)=p(t,T)\left( \delta +(1\color{red}{-}\delta) \mathbb{Q}(\tau > T)\right)$), the spread can be written as $$c(t,T) = - \frac{1}{T-t} \ln \left( \delta +(1\color{red}{-}\delta) \mathbb{Q}(\tau > T)\right), $$ where $\delta$ is the recovery rate and $\mathbb{Q}(\tau > T)$ the risk-neutral probability of the defaultable bond's issuer survival.

I came across the following expression for the "semi-annually compounded" spread, given by $$c(t,T)=2\left[ (\delta +(1-\delta))\mathbb{Q}(\tau > T)) ^{\color{red}{-}\frac{1}{2T}}-1 \right].$$

I don't understand how is this expression derived, could somebody explain it to me? The way I see it, the semi-annually compounded spread should be equal to the difference $y_v-y_p$, where yields $y_v$ and $y_p$ are given implicitly through: $$p(t,T)\left( 1+\frac{y_p}{2} \right)^{2T}=1, $$ and $$v(t,T)\left( 1+\frac{y_v}{2} \right)^{2T}=1. $$ In that case, shouldn't spread be equal to: $$ c(t,T)= 2 \left[ p(t,T)^{-\frac{1}{2T}} \left( 1- \left( \delta +(1+\delta) \mathbb{Q}(\tau > T)\right)^{-\frac{1}{2T}} \right) \right] ? $$

I appreciate any insights on this, many thanks.

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This is all based on how you define the spread. In continuous compounding case, you can define the spread $c(t, T)$ by the formula \begin{align*} v(t, T) = e^{-c(t, T) (T-t)} p(t, T). \end{align*} While, in the semi-annual compounding case, by the formula \begin{align*} v(t, T) = \left(1+\frac{c(t, T)}{2}\right)^{-2 (T-t)} p(t, T). \end{align*}

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