# Given the value of random variable $S_1,$ what is the best prediction $g(S_1)$ for the value of $S_2$?

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?

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

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 $$S_1^{\lambda}e^{\mu_2+\frac{1}{2}\sigma_2^2}$$

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