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I think you are absolutely correct if the hazard rate is deterministic, although I think you are forgetting a discounting factor in your example. But sometimes the hazard rate cannot be assumed to be deterministic (e.g. when pricing CVA and DVA). Here the hazard rate is instead assumed to follow a stochastic process itself, such that $\mathbf{1}_{\{\tau>t\}}$ is the first jump in a Cox process.

It is common to assume that the hazard rate follows a Cox–Ingersoll–Ross process (or extensions to this), which is a mean-reverting square-root diffusion process with SDE $$d \gamma_t=\kappa(\theta-\gamma_t)dt+\sigma\sqrt{\gamma_t}dW_t$$ with Feller-constraint $2\kappa\theta\geq \sigma^2$ to make sure that the origin is inaccessible forcing $\gamma_t>0$ for all $t$.

In general it is assumed that the hazard rate is adapted to the filtration generated by the (default-free) market variables: $\mathscr{F}_t$. Conditional on this information the number of jumps between times $s<t$ is Poisson and the probability of $n$ jumps is thus given by $$\frac{\left(\Gamma_{t}-\Gamma_{s}\right)^{n}}{n!}e^{-\left(\Gamma_{t}-\Gamma_{s}\right)}$$ The probability of zero jumps (no default) is consequently given by $e^{-\left(\Gamma_{t}-\Gamma_{s}\right)}$.

Denote $D(t,T)=e^{-\int_t^Tr_u du}$ such that the risk-neutral valuation of $\xi$ becomes $$\tag{1}V_t=\mathbb{E}\left[D(t,T)\cdot\xi\cdot \mathbf{1}_{\{\tau>T\}}\middle|\mathscr{G}_t\right]$$ Note that $$\tag{2}\mathbb{E}\left[\mathbf{1}_{\{\tau>T\}}\middle|\mathscr{F}_T\vee\mathscr{H}_t\right]=\mathbf{1}_{\{\tau>t\}}e^{-\left(\Gamma_T-\Gamma_t\right)}$$ And note that conditioning on the full filtration would just yield $\mathbb{E}\left[\mathbf{1}_{\{\tau>T\}}\middle| \mathscr{G}_T\right]=\mathbf{1}_{\{\tau>T\}}$, which does not simplify the expression. This means that we can simplify the risk-neutral valuation by utilising the tower property \begin{align*} V_t&=\mathbb{E}\left[D(t,T)\cdot\xi \cdot\mathbb{E}\left[\mathbf{1}_{\{\tau>T\}}\middle|\mathscr{F}_T\vee\mathscr{H}_t\right]\middle|\mathscr{G}_t\right]\\ &=\mathbb{E}\left[D(t,T)\cdot\xi \cdot\mathbf{1}_{\{\tau>t\}}e^{-\left(\Gamma_T-\Gamma_t\right)}\middle|\mathscr{G}_t\right]\\ \tag{3} &=\mathbf{1}_{\{\tau>t\}}\mathbb{E}\left[D(t,T)\cdot\xi \cdot e^{-\left(\Gamma_T-\Gamma_t\right)}\middle|\mathscr{F}_t\right] \end{align*} where the last equality is due to $\mathbf{1}_{\{\tau>t\}}$ being $\mathscr{G}_{t}$-measurable. To simplify further we can assume independence between the discounting factor, the $T$-claim and the hazard rate to obtain $$V_{t}=\mathbf{1}_{\{\tau>t\}}P(t,T)v_{t}\mathbb{E}\left[ e^{-\left(\Gamma_T-\Gamma_t\right)}\middle|\mathscr{F}_t\right]$$ where $P(t,T)$ is a Zero Coupon Bond and the expectation is the probability of not defaulting between $t$ and $T$. This probability can be stripped from spreads on relevant Credit Default Swaps for example.

If we cannot assume independence (for example when there is Wrong Way Risk) then the stochastic dynamics of the hazard rate has to be assumed. So the reason for using two different filtrations is to be able to simplify the expectation when the hazard rate is stochastic.

I think you are absolutely correct if the hazard rate is deterministic, although I think you are forgetting a discounting factor in your example. But sometimes the hazard rate cannot be assumed to be deterministic (e.g. when pricing CVA and DVA). Here the hazard rate is instead assumed to follow a stochastic process itself, such that $\mathbf{1}_{\{\tau>t\}}$ is the first jump in a Cox process.

It is common to assume that the hazard rate follows a Cox–Ingersoll–Ross process (or extensions to this), which is a mean-reverting square-root diffusion process with SDE $$d \gamma_t=\kappa(\theta-\gamma_t)dt+\sigma\sqrt{\gamma_t}dW_t$$ with Feller-constraint $2\kappa\theta\geq \sigma^2$ to make sure that the origin is inaccessible forcing $\gamma_t>0$ for all $t$.

In general it is assumed that the hazard rate is adapted to the filtration generated by the (default-free) market variables: $\mathscr{F}_t$. Conditional on this information the number of jumps between times $s<t$ is Poisson and the probability of $n$ jumps is thus given by $$\frac{\left(\Gamma_{t}-\Gamma_{s}\right)^{n}}{n!}e^{-\left(\Gamma_{t}-\Gamma_{s}\right)}$$ The probability of zero jumps (no default) is consequently given by $e^{-\left(\Gamma_{t}-\Gamma_{s}\right)}$.

Denote $D(t,T)=e^{-\int_t^Tr_u du}$ such that the risk-neutral valuation of $\xi$ becomes $$\tag{1}V_t=\mathbb{E}\left[D(t,T)\cdot\xi\cdot \mathbf{1}_{\{\tau>T\}}\middle|\mathscr{G}_t\right]$$ Note that $$\tag{2}\mathbb{E}\left[\mathbf{1}_{\{\tau>T\}}\middle|\mathscr{F}_T\vee\mathscr{H}_t\right]=\mathbf{1}_{\{\tau>t\}}e^{-\left(\Gamma_T-\Gamma_t\right)}$$ And note that conditioning on the full filtration would just yield $\mathbb{E}\left[\mathbf{1}_{\{\tau>T\}}\middle| \mathscr{G}_T\right]=\mathbf{1}_{\{\tau>T\}}$, which does not simplify the expression. This means that we can simplify the risk-neutral valuation by utilising the tower property \begin{align*} V_t&=\mathbb{E}\left[D(t,T)\cdot\xi \cdot\mathbb{E}\left[\mathbf{1}_{\{\tau>T\}}\middle|\mathscr{F}_T\vee\mathscr{H}_t\right]\middle|\mathscr{G}_t\right]\\ &=\mathbb{E}\left[D(t,T)\cdot\xi \cdot\mathbf{1}_{\{\tau>t\}}e^{-\left(\Gamma_T-\Gamma_t\right)}\middle|\mathscr{G}_t\right]\\ &=\mathbf{1}_{\{\tau>t\}}\mathbb{E}\left[D(t,T)\cdot\xi \cdot e^{-\left(\Gamma_T-\Gamma_t\right)}\middle|\mathscr{G}_t\right]\\ \tag{3} &=\mathbf{1}_{\{\tau>t\}}\mathbb{E}\left[D(t,T)\cdot\xi \cdot e^{-\left(\Gamma_T-\Gamma_t\right)}\middle|\mathscr{F}_t\right] \end{align*} where the second last equality is due to $\mathbf{1}_{\{\tau>t\}}$ being $\mathscr{G}_{t}$-measurable and the last equality is due to the expectation not depending on default information anymore. To simplify further we can assume independence between the discounting factor, the $T$-claim and the hazard rate to obtain $$V_{t}=\mathbf{1}_{\{\tau>t\}}P(t,T)v_{t}\mathbb{E}\left[ e^{-\left(\Gamma_T-\Gamma_t\right)}\middle|\mathscr{F}_t\right]$$ where $P(t,T)$ is a Zero Coupon Bond and the expectation is the probability of not defaulting between $t$ and $T$. This probability can be stripped from spreads on relevant Credit Default Swaps for example.

If we cannot assume independence (for example when there is Wrong Way Risk) then the stochastic dynamics of the hazard rate has to be assumed. So the reason for using two different filtrations is to be able to simplify the expectation when the hazard rate is stochastic.

I think you are absolutely correct if the hazard rate is deterministic, although I think you are forgetting a discounting factor in your example. But sometimes the hazard rate cannot be assumed to be deterministic (e.g. when pricing CVA and DVA). Here the hazard rate is instead assumed to follow a stochastic process itself, such that $\mathbf{1}_{\{\tau>t\}}$ is the first jump in a Cox process.

It is common to assume that the hazard rate follows a Cox–Ingersoll–Ross process (or extensions to this), which is a mean-reverting square-root diffusion process with SDE $$d \gamma_t=\kappa(\theta-\gamma_t)dt+\sigma\sqrt{\gamma_t}dW_t$$ with Feller-constraint $2\kappa\theta\geq \sigma^2$ to make sure that the origin is inaccessible forcing $\gamma_t>0$ for all $t$.

In general it is assumed that the hazard rate is adapted to the filtration generated by the (default-free) market variables: $\mathscr{F}_t$. Conditional on this information the number of jumps between times $s<t$ is Poisson and the probability of $n$ jumps is thus given by $$\frac{\left(\Gamma_{t}-\Gamma_{s}\right)^{n}}{n!}e^{-\left(\Gamma_{t}-\Gamma_{s}\right)}$$ The probability of zero jumps (no default) is consequently given by $e^{-\left(\Gamma_{t}-\Gamma_{s}\right)}$.

Denote $D(t,T)=e^{-\int_t^Tr_u du}$ such that the risk-neutral valuation of $\xi$ becomes $$V_t=\mathbb{E}\left[D(t,T)\cdot\xi\cdot \mathbf{1}_{\{\tau>T\}}\middle|\mathscr{G}_t\right]$$ Note that $$\mathbb{E}\left[\mathbf{1}_{\{\tau>T\}}\middle|\mathscr{F}_T\vee\mathscr{H}_t\right]=\mathbf{1}_{\{\tau>t\}}e^{-\left(\Gamma_T-\Gamma_t\right)}$$ And note that conditioning on the full filtration would just yield $\mathbb{E}\left[\mathbf{1}_{\{\tau>T\}}\middle| \mathscr{G}_T\right]=\mathbf{1}_{\{\tau>T\}}$, which does not simplify the expression. This means that we can simplify the risk-neutral valuation by utilising the tower property \begin{align*} V_t&=\mathbb{E}\left[D(t,T)\cdot\xi \cdot\mathbb{E}\left[\mathbf{1}_{\{\tau>T\}}\middle|\mathscr{F}_T\vee\mathscr{H}_t\right]\middle|\mathscr{G}_t\right]\\ &=\mathbb{E}\left[D(t,T)\cdot\xi \cdot\mathbf{1}_{\{\tau>t\}}e^{-\left(\Gamma_T-\Gamma_t\right)}\middle|\mathscr{G}_t\right]\\ &=\mathbf{1}_{\{\tau>t\}}\mathbb{E}\left[D(t,T)\cdot\xi \cdot e^{-\left(\Gamma_T-\Gamma_t\right)}\middle|\mathscr{F}_t\right] \end{align*} where the last equality is due to $\mathbf{1}_{\{\tau>t\}}$ being $\mathscr{G}_{t}$-measurable. To simplify further we can assume independence between the discounting factor, the $T$-claim and the hazard rate to obtain $$V_{t}=\mathbf{1}_{\{\tau>t\}}P(t,T)v_{t}\mathbb{E}\left[ e^{-\left(\Gamma_T-\Gamma_t\right)}\middle|\mathscr{F}_t\right]$$ where $P(t,T)$ is a Zero Coupon Bond and the expectation is the probability of not defaulting between $t$ and $T$. This probability can be stripped from spreads on relevant Credit Default Swaps for example.

If we cannot assume independence (for example when there is Wrong Way Risk) then the stochastic dynamics of the hazard rate has to be assumed. So the reason for using two different filtrations is to be able to simplify the expectation when the hazard rate is stochastic.

I think you are absolutely correct if the hazard rate is deterministic, although I think you are forgetting a discounting factor in your example. But sometimes the hazard rate cannot be assumed to be deterministic (e.g. when pricing CVA and DVA). Here the hazard rate is instead assumed to follow a stochastic process itself, such that $\mathbf{1}_{\{\tau>t\}}$ is the first jump in a Cox process.

It is common to assume that the hazard rate follows a Cox–Ingersoll–Ross process (or extensions to this), which is a mean-reverting square-root diffusion process with SDE $$d \gamma_t=\kappa(\theta-\gamma_t)dt+\sigma\sqrt{\gamma_t}dW_t$$ with Feller-constraint $2\kappa\theta\geq \sigma^2$ to make sure that the origin is inaccessible forcing $\gamma_t>0$ for all $t$.

In general it is assumed that the hazard rate is adapted to the filtration generated by the (default-free) market variables: $\mathscr{F}_t$. Conditional on this information the number of jumps between times $s<t$ is Poisson and the probability of $n$ jumps is thus given by $$\frac{\left(\Gamma_{t}-\Gamma_{s}\right)^{n}}{n!}e^{-\left(\Gamma_{t}-\Gamma_{s}\right)}$$ The probability of zero jumps (no default) is consequently given by $e^{-\left(\Gamma_{t}-\Gamma_{s}\right)}$.

Denote $D(t,T)=e^{-\int_t^Tr_u du}$ such that the risk-neutral valuation of $\xi$ becomes $$\tag{1}V_t=\mathbb{E}\left[D(t,T)\cdot\xi\cdot \mathbf{1}_{\{\tau>T\}}\middle|\mathscr{G}_t\right]$$ Note that $$\tag{2}\mathbb{E}\left[\mathbf{1}_{\{\tau>T\}}\middle|\mathscr{F}_T\vee\mathscr{H}_t\right]=\mathbf{1}_{\{\tau>t\}}e^{-\left(\Gamma_T-\Gamma_t\right)}$$ And note that conditioning on the full filtration would just yield $\mathbb{E}\left[\mathbf{1}_{\{\tau>T\}}\middle| \mathscr{G}_T\right]=\mathbf{1}_{\{\tau>T\}}$, which does not simplify the expression. This means that we can simplify the risk-neutral valuation by utilising the tower property \begin{align*} V_t&=\mathbb{E}\left[D(t,T)\cdot\xi \cdot\mathbb{E}\left[\mathbf{1}_{\{\tau>T\}}\middle|\mathscr{F}_T\vee\mathscr{H}_t\right]\middle|\mathscr{G}_t\right]\\ &=\mathbb{E}\left[D(t,T)\cdot\xi \cdot\mathbf{1}_{\{\tau>t\}}e^{-\left(\Gamma_T-\Gamma_t\right)}\middle|\mathscr{G}_t\right]\\ \tag{3} &=\mathbf{1}_{\{\tau>t\}}\mathbb{E}\left[D(t,T)\cdot\xi \cdot e^{-\left(\Gamma_T-\Gamma_t\right)}\middle|\mathscr{F}_t\right] \end{align*} where the last equality is due to $\mathbf{1}_{\{\tau>t\}}$ being $\mathscr{G}_{t}$-measurable. To simplify further we can assume independence between the discounting factor, the $T$-claim and the hazard rate to obtain $$V_{t}=\mathbf{1}_{\{\tau>t\}}P(t,T)v_{t}\mathbb{E}\left[ e^{-\left(\Gamma_T-\Gamma_t\right)}\middle|\mathscr{F}_t\right]$$ where $P(t,T)$ is a Zero Coupon Bond and the expectation is the probability of not defaulting between $t$ and $T$. This probability can be stripped from spreads on relevant Credit Default Swaps for example.

If we cannot assume independence (for example when there is Wrong Way Risk) then the stochastic dynamics of the hazard rate has to be assumed. So the reason for using two different filtrations is to be able to simplify the expectation when the hazard rate is stochastic.

I think you are absolutely correct if the hazard rate is deterministic, although I think you are forgetting a discounting factor in your example. But sometimes the hazard rate cannot be assumed to be deterministic (e.g. when pricing CVA and DVA). Here the hazard rate is instead assumed to follow a stochastic process itself, such that $\mathbf{1}_{\{\tau>t\}}$ is the first jump in a Cox process.

It is common to assume that the hazard rate follows a Cox–Ingersoll–Ross process (or extensions to this), which is a mean-reverting square-root diffusion process with SDE $$d \gamma_t=\kappa(\theta-\gamma_t)dt+\sigma\sqrt{\gamma_t}dW_t$$ with Feller-constraint $2\kappa\theta\geq \sigma^2$ to make sure that the origin is inaccessible forcing $\gamma_t>0$ for all $t$.

In general it is assumed that the hazard rate is adapted to the filtration generated by the (default-free) market variables: $\mathscr{F}_t$. Conditional on this information the number of jumps between times $s<t$ is Poisson and the probability of $n$ jumps is thus given by $$\frac{\left(\Gamma_{t}-\Gamma_{s}\right)^{n}}{n!}e^{-\left(\Gamma_{t}-\Gamma_{s}\right)}$$ The probability of zero jumps (no default) is consequently given by $e^{-\left(\Gamma_{t}-\Gamma_{s}\right)}$.

Denote $D(t,T)=e^{-\int_t^Tr_u du}$ such that the risk-neutral valuation of $\xi$ becomes $$V_t=\mathbb{E}\left[D(t,T)\cdot\xi\cdot \mathbf{1}_{\{\tau>T\}}\middle|\mathscr{G}_t\right]$$ Note that $$\mathbb{E}\left[\mathbf{1}_{\{\tau>T\}}\middle|\mathscr{F}_T\vee\mathscr{H}_t\right]=\mathbf{1}_{\{\tau>t\}}e^{-\left(\Gamma_T-\Gamma_t\right)}$$ And note that conditioning on the full filtration would just yield $\mathbb{E}\left[\mathbf{1}_{\{\tau>t\}}\middle| \mathscr{G}_T\right]=\mathbf{1}_{\{\tau>t\}}$, which does not simplify the expression. This means that we can simplify the risk-neutral valuation by utilising the tower property \begin{align*} V_t&=\mathbb{E}\left[D(t,T)\cdot\xi \cdot\mathbb{E}\left[\mathbf{1}_{\{\tau>T\}}\middle|\mathscr{F}_T\vee\mathscr{H}_t\right]\middle|\mathscr{G}_t\right]\\ &=\mathbb{E}\left[D(t,T)\cdot\xi \cdot\mathbf{1}_{\{\tau>t\}}e^{-\left(\Gamma_T-\Gamma_t\right)}\middle|\mathscr{G}_t\right]\\ &=\mathbf{1}_{\{\tau>t\}}\mathbb{E}\left[D(t,T)\cdot\xi \cdot e^{-\left(\Gamma_T-\Gamma_t\right)}\middle|\mathscr{F}_t\right] \end{align*} where the last equality is due to $\mathbf{1}_{\{\tau>t\}}$ being $\mathscr{G}_{t}$-measurable. To simplify further we can assume independence between the discounting factor, the $T$-claim and the hazard rate to obtain $$V_{t}=\mathbf{1}_{\{\tau>t\}}P(t,T)v_{t}\mathbb{E}\left[ e^{-\left(\Gamma_T-\Gamma_t\right)}\middle|\mathscr{F}_t\right]$$ where $P(t,T)$ is a Zero Coupon Bond and the expectation is the probability of not defaulting between $t$ and $T$. This probability can be stripped from spreads on relevant Credit Default Swaps for example.

If we cannot assume independence (for example when there is Wrong Way Risk) then the stochastic dynamics of the hazard rate has to be assumed. So the reason for using two different filtrations is to be able to simplify the expectation when the hazard rate is stochastic.

I think you are absolutely correct if the hazard rate is deterministic, although I think you are forgetting a discounting factor in your example. But sometimes the hazard rate cannot be assumed to be deterministic (e.g. when pricing CVA and DVA). Here the hazard rate is instead assumed to follow a stochastic process itself, such that $\mathbf{1}_{\{\tau>t\}}$ is the first jump in a Cox process.

It is common to assume that the hazard rate follows a Cox–Ingersoll–Ross process (or extensions to this), which is a mean-reverting square-root diffusion process with SDE $$d \gamma_t=\kappa(\theta-\gamma_t)dt+\sigma\sqrt{\gamma_t}dW_t$$ with Feller-constraint $2\kappa\theta\geq \sigma^2$ to make sure that the origin is inaccessible forcing $\gamma_t>0$ for all $t$.

In general it is assumed that the hazard rate is adapted to the filtration generated by the (default-free) market variables: $\mathscr{F}_t$. Conditional on this information the number of jumps between times $s<t$ is Poisson and the probability of $n$ jumps is thus given by $$\frac{\left(\Gamma_{t}-\Gamma_{s}\right)^{n}}{n!}e^{-\left(\Gamma_{t}-\Gamma_{s}\right)}$$ The probability of zero jumps (no default) is consequently given by $e^{-\left(\Gamma_{t}-\Gamma_{s}\right)}$.

Denote $D(t,T)=e^{-\int_t^Tr_u du}$ such that the risk-neutral valuation of $\xi$ becomes $$V_t=\mathbb{E}\left[D(t,T)\cdot\xi\cdot \mathbf{1}_{\{\tau>T\}}\middle|\mathscr{G}_t\right]$$ Note that $$\mathbb{E}\left[\mathbf{1}_{\{\tau>T\}}\middle|\mathscr{F}_T\vee\mathscr{H}_t\right]=\mathbf{1}_{\{\tau>t\}}e^{-\left(\Gamma_T-\Gamma_t\right)}$$ And note that conditioning on the full filtration would just yield $\mathbb{E}\left[\mathbf{1}_{\{\tau>T\}}\middle| \mathscr{G}_T\right]=\mathbf{1}_{\{\tau>T\}}$, which does not simplify the expression. This means that we can simplify the risk-neutral valuation by utilising the tower property \begin{align*} V_t&=\mathbb{E}\left[D(t,T)\cdot\xi \cdot\mathbb{E}\left[\mathbf{1}_{\{\tau>T\}}\middle|\mathscr{F}_T\vee\mathscr{H}_t\right]\middle|\mathscr{G}_t\right]\\ &=\mathbb{E}\left[D(t,T)\cdot\xi \cdot\mathbf{1}_{\{\tau>t\}}e^{-\left(\Gamma_T-\Gamma_t\right)}\middle|\mathscr{G}_t\right]\\ &=\mathbf{1}_{\{\tau>t\}}\mathbb{E}\left[D(t,T)\cdot\xi \cdot e^{-\left(\Gamma_T-\Gamma_t\right)}\middle|\mathscr{F}_t\right] \end{align*} where the last equality is due to $\mathbf{1}_{\{\tau>t\}}$ being $\mathscr{G}_{t}$-measurable. To simplify further we can assume independence between the discounting factor, the $T$-claim and the hazard rate to obtain $$V_{t}=\mathbf{1}_{\{\tau>t\}}P(t,T)v_{t}\mathbb{E}\left[ e^{-\left(\Gamma_T-\Gamma_t\right)}\middle|\mathscr{F}_t\right]$$ where $P(t,T)$ is a Zero Coupon Bond and the expectation is the probability of not defaulting between $t$ and $T$. This probability can be stripped from spreads on relevant Credit Default Swaps for example.

If we cannot assume independence (for example when there is Wrong Way Risk) then the stochastic dynamics of the hazard rate has to be assumed. So the reason for using two different filtrations is to be able to simplify the expectation when the hazard rate is stochastic.