Show that $Y_t \int_0^t \frac{1}{Y_s} ds$ and $\int_0^t Y_s ds$ are equal in distribution

Question

Let $B=(B_t)_{t \geq0}$ be a standard Brownian motion and let the two stochastic processes $X=(X_t)_{t \geq0}$ and $Y=(Y_t)_{t \geq0}$ solve the following two stochastic differential equations

$$dX_t = (1+3X_t)dt+ (1+5X_t)dB_t$$

and

$$dY_t = 3Y_tdt+ 5Y_tdB_t$$

respectively. Also we have $X_0 = Y_0 =1$. The task is to show that

$$X_t- Y_t(1+\int_0^t\frac{1}{Y_s}dB_s) \stackrel{d}{=} -4\int_0^tY_s ds.$$

I have already shown using Ito's lemma that

$$X_t = Y_t(1+\int_0^t\frac{1}{Y_s}dB_s -4\int_0^t \frac{1}{Y_s}ds)$$

thus to finish the problem I need to show that

$$Y_t \int_0^t \frac{1}{Y_s} ds \stackrel{d}{=} \int_0^t Y_s ds.$$

I'm not really sure how to go about doing this? The wording is actually equal in law but I assume this means equal in distribution?

Edit:

I have shown that $$Y_s \sim \text{Lognormal}(-\frac{19s}{2}, 25s)$$ and $$\frac{Y_t}{Y_s} \sim \text{Lognormal}(-\frac{19(t-s)}{2}, 25(t-s)).$$ Is this sufficient?

Answer 1

The equality in law $$ \int_0^t \frac{Y_t}{Y_s} ds \stackrel{d}{=} \int_0^t Y_s ds $$ is indeed true.

Just note that (using simple calculus) the rhs side is equal to $\int_0^t Y_{t-s} ds$

Now, the processes $s \mapsto \frac{Y_t}{Y_s} $ and $s \mapsto Y_{t-s} $ are equal in law (do you see why?), so their time integrals will also be equal in law.