Let $(B_t)_{t \in [0,T]}$ be standard brownian motion, and let $(Y_t)_{t \in [0,T]}$ be a stochastic process in $(\Omega, \mathscr F, \mathbb P)$.

Without using the general linear SDE formula, solve the SDE:

$$dY_t = tY_tdt + e^{t^2/2}dB_t$$ by considering $$Y_t = a(t) \left(Y_0 + \int_0^t b(s) dB_s \right)$$

with initial condition $Y_0 = \alpha \in \mathbb R$, where $a(t)$ and $b(t)$ are differential functions that are not random.

Hint: Consider

$$Y_t = a(t) \left(Y_0 +\int_{0}^{t} b(s) dB_s \right)$$

What I tried for solving:

$$dY_t = a'(t)Y_0 dt + a'(t) \ dt \ b(t) \ dB_t + a(t)b(t)dB_t + a'(t) dt \int_0^t b(s) dB_s$$

$$= a'(t)Y_0dt + a(t)b(t)dB_t + a'(t) dt \int_0^t b(s) dB_s$$

$$= a'(t) \alpha dt + a(t)b(t)dB_t + a'(t) dt \int_0^t b(s) dB_s$$

$$= Y_t\frac{a'(t)}{a(t)}dt + a(t)b(t)dB_t$$


$$Y_t\frac{a'(t)}{a(t)}dt + a(t)b(t)dB_t = tY_tdt + e^{t^2/2}dB_t,$$

we get that

$$a(t) = Ae^{t^2/2}$$

where $A = \pm e^C$ where $C \in \mathbb R$ and

$$b(t) = 1/A$$


$$Y_t = \alpha Ae^{t^2/2} + e^{t^2/2}B_t$$

Is that right? I find that A kind of weird. I think it's supposed to cancel out.

What I tried for proving uniqueness:

  1. Show that $E[Y_0^2] < \infty$:

$$E[Y_0^2] = E[\alpha^2] = \alpha^2 < \infty$$

  1. Show $\exists K \in \mathbb R$ s.t.

$|tx| + |e^{t^2/2}| \leq K(|x|+1)$

$|tx-ty| \leq K(|x-y|)$:

We have $$|t| = t \le T \to e^{t^2/2} \le e^{T^2/2}$$

Choose $$K = e^{T^2/2}$$

Is that right?

  • 1
    $\begingroup$ One line proof: $$dY_t = tY_tdt + e^{t^2/2}dB_t\iff d(e^{-t^2/2}Y_t)=dB_t$$ $\endgroup$
    – Did
    Nov 21, 2015 at 16:30

1 Answer 1


Note that

$$Y_t = \alpha A e^{t^2/2} + e^{t^2/2} B_t$$

has to satisfy the initial condition $Y_0 = \alpha$. Thus,

$$Y_0 = \alpha A \stackrel{!}{=} \alpha,$$

i.e. $A=1$. Consequently,

$$Y_t = e^{t^2/2} (\alpha+B_t).$$

If you want to check whether this is indeed a solution to the given SDE, then just apply Itô's formula with

$$f(t,x) := e^{t^2/2} (\alpha+x),$$


$$Y_t -Y_0 = \int_0^t e^{s^2/2} \, dB_s + \int_0^t s e^{s^2/2} (\alpha+B_s) \, ds = \int_0^t e^{s^2/2} \, dB_s + \int_0^t s Y_s \, ds.$$

This shows that $(Y_t)_{t \geq 0}$ solves the given SDE.

(And, yeah, the constants for the proof of the uniqueness are okay.)

  • $\begingroup$ Thanks saz! Do I have to use do that to check? Why can't I just take the differential like I did following the hint? $\endgroup$
    – BCLC
    Nov 21, 2015 at 17:08
  • 1
    $\begingroup$ @BCLC In this particular case, you don't have to check it. However, if you are not sure/convinced whether your solution is correct, checking it using Itô's formula is always a good idea. $\endgroup$
    – saz
    Nov 21, 2015 at 17:12

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