Couldn't understand decomposition inside derivation of option valuation using Martingale method In the derivation of option valuation using Martingale method in continuous time framework of my book named "Brownian Motion Calculus by Ubbo F Wiersema" I have faced some issue. I add the derivation below:
In the continuous time setting, start from the standard SDE for the stock price, $$\frac{dS_t}{S_t}=\mu dt+\sigma dB_t$$
Introduce the discounted stock price, $$S^{\star}_t\stackrel{\text{def}}{=}\frac{S_t}{e^{rt}}$$
Ito's formula for $S^{\star}_t$ as a function of $t$ and $S$ gives, $$\frac{dS^{\star}_t}{S^{\star}_t}=(\mu-r)dt+\sigma dB_t=\sigma\left[ \frac{\mu-r}{\sigma}dt+dB_t \right]=\sigma\left[ \phi dt+dB_t \right]$$
The probability density of $B_t$ at $B_t=x$ is, $$\frac{1}{\sqrt t \sqrt{2\pi}}\exp\left[-\frac12\left(\frac{x}{\sqrt t}\right)^2\right]$$
This can be decomposed into the product of two terms,

$$\frac{1}{\sqrt t \sqrt{2\pi}}\exp\left[-\frac12\left(\frac{\phi t+x}{\sqrt t}\right)^2\right]\exp\left[\frac12 \phi^2 t+\phi x\right]$$

With $y \stackrel{\text { def }}{=} \varphi t+x$ the first term can be written as
$$
\frac{1}{\sqrt{t} \sqrt{2 \pi}} \exp \left[-\frac{1}{2}\left(\frac{y}{\sqrt{t}}\right)^{2}\right]
$$
which is the probability density of another Brownian motion, say $\widehat{B}(t)$, at $\widehat{B}(t)=y$. It defines $\widehat{B}(t) \stackrel{\text { def }}{=} \varphi t+B(t)$, so $\widehat{d} \widehat{B}(t)=\varphi d t+d B(t)$. Substituting the latter into the SDE for $S^{\star}$ gives
$$
\frac{d S^{\star}(t)}{S^{\star}(t)}=\sigma d \widehat{B}(t) \quad \text { and } \quad \frac{d S(t)}{S(t)}=r d t+\sigma d \widehat{B}(t)
$$
This says that under the probability distribution of Brownian motion $\widehat{B}(t)$,
$S^{\star}$ is a martingale.
I couldn't understand how they bring the decomposition like this? Another thing is, "Why we need to sustain martingale?". Does it mean we couldn't predict the future regardless of all prior knowledge, as it introduce arbitrage?

@Ali, give an answer to show the decomposition.
$$
\exp\left[-\frac{1}{2}\left(
\frac{\phi t + x}{\sqrt{t}}\right)^{2}\right]=
\exp\left[-\frac{1}{2}\left(
\phi^{2}t+2\phi x + x^{2}/t\right)\right]=
\exp\left[-\frac{1}{2}
\phi^{2}t-\phi x\right]
\exp\left[-\frac{1}{2t}x^{2}\right]
$$
But, still, I couldn't get any intuition of that decomposition. Like, I want to know why this decomposition was introduced, and its meaning in the context of option valuation.
Does this related with Girsanov transformation?
 A: @Ali already showed the algebraical manipulation. Let me give you some explanation. You have to understand the significance of martingales in Option Pricing. Fundamental facts of financial derivatives is: $$\text{No Arbitrage} \iff \exists \text{(a Martingale measure)}$$
For more details, have a look at here.
When you have, $$\frac{dS^{\star}_t}{S^\star_t}=(\mu-r)dt+\sigma dB_t$$
means the price of that option written on a security follows a geometric Brownian motion which is a function of the current price of the security, and not its price history. But after the transformation, $$\frac{dS^{\star}_t}{S^\star_t}=\sigma d \widehat{B}(t)$$
we refer to it as driftless geometric Brownian motion (aka martingale). For more details, have a look at here. Much of Asset Pricing Theory characterizes fair value for risky securities in terms of martingales.
[Durrett proves Girsanov’s theorem in the generality needed for
empirically based option pricing: One starts with any Martingale $x(t)$, adds an $(x,t)$ dependent
drift term $R(x,t)$, and then constructs a new Martingale. The new
Martingale is not a Wiener process unless -

*

*(a) the old process is Wiener,

*(b) the drift is not a function of a random variable $x(t)$.

The sole exception to this rule is the lognormal SDE $dx=\mu x dt+\sigma^2 x dB$ with $\sigma$ constant, which is trivially Wiener by the simple coordinate transformation $y=\ln x$.
A: $$
\exp\left[-\frac{1}{2}\left(
\frac{\phi t + x}{\sqrt{t}}\right)^{2}\right]=
\exp\left[-\frac{1}{2}\left(
\phi^{2}t+2\phi x + x^{2}/t\right)\right]=
\exp\left[-\frac{1}{2}
\phi^{2}t-\phi x\right]
\exp\left[-\frac{1}{2t}x^{2}\right]
$$
