Heston model passages

Question

I'm struggling with a passage of the demonstration of Heston model. I have to prove that $E^{\mathbb{Q}}[S_T \mathbb{I}_{S_T \geq K}]e^{-rT}=S_0E^{\mathbb{Q}^S}[\mathbb{I}_{S_T \geq K}]$ using Radon-Nikodym derivative $\frac{d\mathbb{Q}^S}{d\mathbb{Q}}=\frac{S_T}{B_TS_0}$. I know that:

• $B_T:=e^{-\int_{0}^{T}rds}=e^{-rT}$ for $r$ deterministic risk-free rate.
• $B_0:=1$
• $S_0:=S_Te^{-rT}$

Can you help me? Thanks in advance!

Answer 1

Note that $B_T = B_0e^{rT}$, not $e^{-rT}$!

What I think your professor has done is the following (I will set $X = \mathbb{I}_{S_T\geq K}$ for simplicity):

\begin{align} e^{-rT}\text{E}_Q\left[S_T X\right] &= S_0 e^{-rT}\text{E}_Q\left[\frac{1}{S_0}S_T X\right]\\ &= S_0\text{E}_Q\left[\frac{S_T}{S_0} e^{-rT} X\right]\\ \end{align}

Now, using the fact that $B_T = e^{rT}$:

\begin{align} &= S_0\text{E}_Q\left[\frac{S_T}{S_0} \frac{1}{B_T} X\right]\\ &= S_0\text{E}_Q\left[\frac{S_T/ S_0}{B_T}X\right]\\ &= S_0\text{E}_Q\left[\frac{S_T/ S_0}{B_T / B_0}X\right] \end{align}

Now, we use the Radon-Nikodym Derivative in the following way:

$$ \text{E}_{Q}[X] = \text{E}_{P}\left[X \frac{dQ}{dP}\right] $$

So by the same Logic:

\begin{align} S_0\text{E}_Q\left[\frac{S_T/ S_0}{B_T / B_0}X\right] &= S_0\text{E}_{Q_s}\left[\frac{S_T/ S_0}{B_T / B_0}X \cdot \frac{dQ}{dQ_s}\right] \\ \end{align}

Now, assuming the measures $Q$ and $Q_s$ are equivalent, we have

$$ \frac{dQ}{dQ_s} = \frac{1}{dQ_s / dQ} $$

which means

$$ \frac{dQ}{dQ_s} = \frac{B_T S_0}{S_T} $$

we then have

\begin{align} S_0\text{E}_{Q_s}\left[\frac{S_T/ S_0}{B_T / B_0}X \cdot \frac{dQ}{dQ_s}\right] &= S_0\text{E}_{Q_s}\left[\frac{S_T/ S_0}{B_T / B_0} \frac{B_T S_0}{S_T} X\right]\\ &= S_0\text{E}_{Q_s}\left[\frac{1}{B_0}X\right]\\ &= S_0\text{E}_{Q_s}\left[X\right]\\ \end{align}

Since we have set $X = \mathbb{I}_{S_T\geq K}$, we end up with

$$ S_0\text{E}_{Q_s}\left[\mathbb{I}_{S_T\geq K}\right],\\ $$

as required.

Answer 2

This is the proof of professor.

I tried to write: $E^{\mathbb{Q}}[S_T \mathbb{I}_{S_T \geq K}]e^{-rT}=E^{\mathbb{Q}^S}[\frac{S_T}{B_TS_0}\mathbb{I}_{S_T \geq K} \frac{d \mathbb{Q}^S}{d \mathbb{Q}}]e^{-rT}=E^{\mathbb{Q}^S}[\frac{S_T \cdot 1}{B_TS_0}\mathbb{I}_{S_T \geq K} \frac{d \mathbb{Q}^S}{d \mathbb{Q}}]e^{-rT}=E^{\mathbb{Q}^S}[\frac{S_T \cdot B_0}{B_TS_0}\mathbb{I}_{S_T \geq K} \frac{d \mathbb{Q}^S}{d \mathbb{Q}}]e^{-rT}$

Then I know that $\frac{S_T}{S_0}=e^{-rT}$ and $\frac{B_0}{B_T}=\frac{1}{e^{\int_{0}^{T}rds}}=e^{-rT}$, but I can't delete $e^{-rT}$. Professor, then, multiplies and divides for $S_0$...