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Suppose $$ S_1 = \exp(X_1), \quad \quad X_1 \sim N(\mu_1, \sigma_1^2) $$

$$ S_2 = \exp(\lambda X_1 + X_2), \quad \quad X_2 \sim N(\mu_2, \sigma_2^2). $$ Assume $X_1$ and $X_2$ are independent and $\lambda \in \mathbb{R}$, and $\mu_1, \mu_2, \sigma_1, \sigma_2, \lambda$ are all known.

Given $S_1$, what is the best prediction we can make for the value of $S_2$, written as a function $g(S_1)$?

Here was my approach so far:

We want to find $\mathbb{E}(S_2 | S_1).$ \begin{align*} \mathbb{E}(S_2 | S_1) &= \mathbb{E}\lbrack \exp(\lambda X_1+ X_2) | \exp(X_1) \rbrack \\ &= \mathbb{E} \lbrack e^{\lambda X_1} \cdot e^{X_2}|e^{X_1} \rbrack \\ &= e^{\lambda X_1} \mathbb{E}\lbrack e^{X_2}|e^{X_1}\rbrack \end{align*} since we can "take out what is known" from within the conditional expectation. Now, this is what I'm unsure about. I believe the independence of $X_1$ and $X_2$ would imply the independence of $e^{X_1}$ and $e^{X_2}$. If this were true, then $$ e^{\lambda X_1} \mathbb{E}\lbrack e^{X_2}|e^{X_1}\rbrack = e^{\lambda X_1} \cdot e^{X_2} = e^{X_2}\cdot S_1^\lambda. $$ Like I said, I'm not sure about the independence of $e^{X_1}$ and $e^{X_2}$. If this is not true, why?

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Since $X_1$ and $X_2$ are independent, it is true that $e^{X_1}$ and $e^{X_2}$ are independent. However, this implies $$ \mathbb{E}[e^{X_2}\mid e^{X_1}]=\mathbb{E}[e^{X_2}] $$ which is not what you have written.

Following up, this leads to $$ \mathbb{E}[S_2\mid S_1] = e^{\lambda X_1}\mathbb{E}[e^{X_2}] = e^{\lambda X_1} e^{\mu_2+\frac{1}{2}\sigma_2^2} = \boxed{S_1^\lambda \cdot e^{\mu_2+\frac{1}{2}\sigma_2^2}}\,. $$

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You forgot the expectation term, we have

$$\mathbb{E}\lbrack e^{X_2}|e^{X_1}]=\mathbb{E}\lbrack e^{X_2}]$$

You should then get $S_1^{\lambda}e^{\mu_2+\frac{1}{2}\sigma_2^2}$

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    $\begingroup$ $S_1^\lambda \cdot e^{\mu_2+\frac{1}{2}\sigma_2^2}$, rather. $\endgroup$ – Clement C. Dec 5 '19 at 18:50
  • $\begingroup$ Ah yes I see. Thank you. $\endgroup$ – dove Dec 5 '19 at 18:50
  • $\begingroup$ Exactly Clement C. $\endgroup$ – Canardini Dec 5 '19 at 18:51

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