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Let's say I have two processes $X_t =X_0 \exp((a-\frac{1}{2}\sigma_X^2)t +\sigma_X dW_t^1)$ and $Y_t=Y_0 \exp((b-\frac{1}{2}\sigma_Y^2)t +\sigma_Y dW_t^2)$ and I then multiply them together (like converting a foreign asset into domestic currency). I arrive at

$X_tY_t = X_0Y_0 \exp((a+b-\frac{1}{2}\sigma_X^2-\frac{1}{2}\sigma_Y^2)t +\sigma_X dW_t^1+\sigma_Y dW_t^2)$

If I then assume the $W_t$ have a correlation of $\rho$, how would I obtain the expectation of $X_tY_t$. The thing that is troubling me is because I have 2 Brownian motions and I'm not sure how to compute from there.

Edit: I would like to make this question a little bit more general. What would I do if I wanted to compute $E[X_t Y_s]$, where say $s < t$. So something to do with independent increments would play a factor.

Jim
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1 Answers1

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Just use the fact that $$ \sigma_X W_t^1 + \sigma_Y W_t^2 = \sqrt{ \sigma_X^2 + \sigma_Y^2 + 2\rho\sigma_X\sigma_Y } W_t $$ holds in probability assuming that $W_t^1$ and $W_t^2$ are 2 correlated Brownian motions with $$ d\langle W_t^1, W_t^2 \rangle_t = \rho dt $$ and $W_t$ is a new standard Brownian motion defined over the same probability space.

Simply put, just replace your sum of two correlated Gaussians (LHS above) by a single Gaussian (RHS above) exhibitting the exact same statistical properties (for a Gaussian identical mean/variance is enough). Doing so, you can now use the formulas you are accustomed to.

Applying this shows $$ X_t Y_t = X_0Y_0 \exp((a+b-\frac{1}{2}\sigma_X^2-\frac{1}{2}\sigma_Y^2)t +\sqrt{ \sigma_X^2 + \sigma_Y^2 + 2\rho\sigma_X\sigma_Y } W_t) $$ is lognormally distributed with mean $$ \mu = \ln(X_0Y_0)+(a+b-\frac{1}{2}\sigma_X^2-\frac{1}{2}\sigma_Y^2)t $$ and variance $$ \sigma^2 = (\sigma_X^2 + \sigma_Y^2 + 2\rho\sigma_X\sigma_Y)t $$

hence applying the usual formula for the mean of a lognormally distributed variable $$ E[X_t Y_t] = e^{\mu + \sigma^2/2} $$ is a function of $\rho$.


[Edit]

I think you are completely mixing up two different problems (here two different probability measures).

I understand from your comment that you define the dynamics of the FOR/DOM instantaneous exchange rate $X_t$ (i.e. $X_t = x$ meaning that, at time $t$, 1 unit of foreign currency = x units of domestic currency) under the foreign risk-neutral measure $\mathbb{Q}^f$ (or rather the probability space $(\Omega,\mathcal{F},\mathbb{Q}^f)$ along with the price process of an equity underlying denominated in the foreign currency.

In that case, under the domestic risk-neutral measure $\mathbb{Q}^d$ (or rather the probability space $(\Omega,\mathcal{F},\mathbb{Q}^d)$), one should have, by absence of arbitrage opportunities and assuming market completeness: $$ \frac{Y_t X_t}{B_t^d} \text{ is a } \mathbb{Q}^d \text{- martingale} $$

with $B^d_t$ representing the time-$t$ value of a risk-free money market account in the domestic economy in which 1 unit of currency has been invested at $t=0$. Using the martingale property it then entails that: $$ \frac{Y_0 X_0}{B_0^d} = Y_0 X_0 = E^{\mathbb{Q}^d} \left[ \frac{Y_t X_t}{B_t^d} \vert \mathcal{F}_0 \right] $$ and further assuming deterministic rates: $$ E ^{\mathbb{Q}^d} \left[ X_t Y_t \vert \mathcal{F}_0 \right] = e^{-r_d t} X_0 Y_0 $$ which is independent of $\rho$.

This is known as a compo adjustment in deriatives lingo.

That being said, under the measure $\mathbb{Q}^f$ where you have originally defined $X_t$ and $Y_t$, the expectation $E^{\mathbb{Q}^f} [ Y_t X_t ]$ does dependent on $\rho$ as hinted above.

More info on the quanto/compo change of measure technique here

Quantuple
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  • Thanks for that, but I guess I might be having trouble applying this to my problem. My problem has $X_t =X_0\exp((r_d-r_f-\frac{1}{2}\sigma_X^2)t+\sigma_X W_t^1)$ and $Y_t =Y_0\exp((r_f-\frac{1}{2}\sigma_Y^2)t+\sigma_Y W_t^2)$. What would be the expectation of $X_tY_t$ because my answer involves $\rho$, when I know for certain that it should not. – Jim May 03 '16 at 11:36
  • @Jim, you are actually mixing up two different problems, please see my edits – Quantuple May 03 '16 at 12:52
  • The relationship you are trying to show is only true for a specific situation known as compo adjustment (has to do with quanto changes of measures), I would advise to change the title of your question to reflect that, for it is an interesting question, not asked in the right way. – Quantuple May 03 '16 at 12:54
  • Great explanation of my two minds. Would you be able to show how you arrive at the expectation for the compo adjustment (as you call it). Do I need to use Girsanov? – Jim May 03 '16 at 13:06
  • Do you agree that if you manage to show that $Y_t X_t / B_t^d$ is a $\mathbb{Q}^d$-martingale then you are done? – Quantuple May 03 '16 at 13:07
  • Yes, I agree with that. But how would one go from the $Q^f$ measure to the $Q^d$ measure. – Jim May 03 '16 at 13:11
  • Well this martingale property is simply a consequence of the fundamental theorems of asset pricing as some call them: in the absence of arbitrage opportunities (+ complete market) there exists a unique risk-neutral measure in which all tradable asset prices are martingales when the risk-free money market account is used as numéraire. Here, we look at the domestic economy. The foreign asset $Y_t$ is not directly tradable in that economy (since denominated in foreign currency units), so you first need to convert $Y_t$ to domestic units $Y_t X_t$ before applying the numéraire $B^d_t$ – Quantuple May 03 '16 at 13:13
  • Ah yes, you do need Girnasov in that case, I will edit my post give me 2 secs. – Quantuple May 03 '16 at 13:14
  • @Jim, hope this clears everything. – Quantuple May 03 '16 at 13:32
  • Would you be able to quickly show the Girsanov transformation. I've never worked with these measures before so it's a little confusing for me. Other than that, this has been excellent – Jim May 03 '16 at 13:44
  • @Jim Sorry but this has already gone too far out of topic, IMO your original question has already been answered. It is better to keep the 1 question/1 answer format, this is not a forum! So I would be pleased to help you; but just ask a new question publicly. – Quantuple May 03 '16 at 13:59
  • I have added in an edit to try and generalise this result. Would you be able to help with that. – Jim May 04 '16 at 14:43
  • @Jim, I'm sure everyone here would be glad to help. But as we already discussed, the Stack Exchange format is 1 question / 1 answer. So you better ask this in a new, separate, question – Quantuple May 04 '16 at 14:51
  • Oh I thought that since it was just a generalisation, I wouldn't need a new question. – Jim May 04 '16 at 15:02