Solution for SDE: $dF_t= \beta_t\left(F_t - \alpha\right)dW_t$

I am trying to derive the solution for the following stochastic differential equation, but I must be doing something wrong in my calculations because I can't arrive to the correct solution.

The SDE in question is the following: $$dF_t = \beta_t\left(F_t - \alpha\right)dW_t$$

where $W$ is a standard Brownian Motion started at $0$, $\alpha$ is a constant and $\beta(t)$ is a deterministic process. I know from the literature that the solution for this SDE for $0 < t < T$ can be explicitly written as

$$F_T = \alpha + (F_t - \alpha) e^{-\frac{1}{2}\int_t^T\beta^2_udu + \int_t^T\beta_udW_u}$$

Surely I am not applying Ito's formula where I should be?

Bonus question: What is the impact of applying a shift to the distribution generated by the above equation? When computing probabilities for different values, how big will be the differences in the case where $\alpha = 0$ and $\alpha \neq 0$? Intuitively I would like to think that the results would be the same regardless of the shift applied.

I will first make a few simplifications. Define $G_t := F_t - \alpha$. Then of course $dG_t = dF_t$. So then we are solving for $dG_t = \beta_tG_tdW_t$. Since $t\mapsto\beta_t$ is deterministic, this is still geometric Brownian motion. If $t\mapsto\beta_t$ is Lipschitz, which you would assume to guarantee uniqueness, then $G_t$ has the Markov property. So then, we can look at the solution $$G_t = G_0e^{-\frac{1}{2}\int_0^t\beta_s^2\,ds + \int_0^t\beta_s\,dW_s}$$ instead of arbitrary $0 \leq t\leq T$.

Define the function $f_t = \int_0^t\beta_s^2\,ds$ and the process $X_t = \int_0^t\beta_s\,dW_s$. Hence, $G_t = G_0e^{-\frac{1}{2}f_t + X_t}$. Note that $dX_t = \beta_tdW_t$ and $d[X,X]_t = \beta^2_tdt$. Since you wanted a full analytical solution, I will be a bit pedantic and treat $G_0$ as another stochastic process (constant over time but still random), i.e. $Y_t = G_0$ for all $t \geq 0$. Then of course $dY_t = 0$ and $d[Y,Y]_t = 0$. I hope it is also clear to you that $d[X,Y]_t = 0$ since all the paths of $Y$ are constant.

Now consider the function $r : (t,x,y) \mapsto ye^{-\frac{1}{2}f_t + x}$. You can see that $G_t = r(t,X_t,Y_t)$. This function is infinitely differentiable in all its arguments so we have more than what we need to apply Ito's lemma. I will use subscripts to denote partial derivatives of $r$ below.

$$dG_t = r_t(t,X_t,Y_t)dt + r_x(t,X_t,Y_t)dX_t + r_y(t,X_t,Y_t)dY_t + \frac{1}{2}r_{xx}(t,X_t,Y_t)d[X,X]_t + \frac{1}{2}r_{yy}(t,X_t,Y_t)d[Y,Y]_t + r_{xy}(t,X_t,Y_t)d[X,Y]_t$$

The last two terms and the $dY$ term drop due to the considerations above. $r_t = -\frac{1}{2}\frac{df}{dt}r = -\frac{1}{2}\beta^2_tr$, $r_x = r$ and $r_{xx} = r$. Putting all these together,

$$dG_t = -\frac{1}{2}\beta^2_tG_tdt + G_t \beta_tdW_t + \frac{1}{2}G_t\beta^2_tdt = G_t \beta_tdW_t$$

So then the given solution satisfies the SDE. For the other question we need to look at

$$P\{F_T \leq x\} = P\{\alpha + (F_0-\alpha)e^{-\frac{1}{2}\int_0^t\beta_u^2\,du + \int_0^t\beta_u\,dW_u} \leq x\}$$ $$P\{F_T \leq x\} = P\{e^{-\frac{1}{2}\int_0^t\beta_u^2\,du + \int_0^t\beta_u\,dW_u} \leq \frac{x-\alpha}{F_0 -\alpha}\}$$ $$P\{F_T \leq x\} = P\{-\frac{1}{2}\int_0^t\beta_u^2\,du + \int_0^t\beta_u\,dW_u \leq \log\left(\frac{x-\alpha}{F_0 -\alpha}\right)\}$$ $$P\{F_T \leq x\} = P\{\int_0^t\beta_u\,dW_u \leq \log\left(\frac{x-\alpha}{F_0 -\alpha}\right) + \frac{1}{2}\int_0^t\beta_u^2\,du\}$$

Note that $\int_0^t\beta_u\,dW_u \sim \mathsf{N}(0,\int_0^t\beta_u^2\,du)$. Then we can write $$P\{F_T \leq x\} = \Phi\left(\frac{\log\left(\frac{x-\alpha}{F_0 -\alpha}\right) + \frac{1}{2}\int_0^t\beta_u^2\,du}{\sqrt{\int_0^t\beta_u^2\,du}}\right)$$ Here $\Phi$ is the distribution function of a standard normal random variable. The sensitivity of this expression with respect to $\alpha$ you can judge for yourself by looking at the derivatives or using numerical methods.

One last remark: In the second part I conveniently assumed that $F_0$ is a constant. $F_0$ can be a random variable itself. Note that for $F_0$ to be measurable with respect to $\mathcal{F}_0$ it would have to be independent of $W$. Then, all you need to specify is the marginal distribution of $F_0$ to evaluate the probability $P\{F_T \leq x\}$ (rather than the joint distribution of $\int_0^t\beta_u\,dW$ and $F_0$).

• Through answer indeed. Regarding the bonus question, I am asking whether $\mathbb{P}(F \leq x)$ returns the same result regardless of $\alpha$ being $\neq 0$ or $= 0$.