Can it be that $W(t)$ induces $\mathbb{P}$ (and therefore $\mathbb{P}$ is defined by the density of $W(t)$): and then, you have $X(t):=W(t)−rt$ (and therefore $X(t)$ has a different density) under the same measure $\mathbb{P}$ ? Wouldn't $X(t)$ define its own measure via its own density?

Specifically, we can define $\mathbb{P}$ for any $A:a \epsilon \mathbb{R}$ as follows: $\mathbb{P}(A):=\int^{a}_{-\infty}f_{W_t}(h)dh$

We could then define $\mathbb{Q}$ for any $A:a \epsilon \mathbb{R}$ as follows: $\mathbb{Q}(A):=\int^{a}_{-\infty}f_{X_t}(h)dh$.

In the above, $f_{W_t}()$ is Normal density with variance $t$, whilst $f_{X_t}()$ is Normal density with variance $t$ and mean $-rt$.

The R-N Derivative to get from $\mathbb{P}$ to $\mathbb{Q}$ would then be $\frac{\mathrm{d}\mathbb{Q}}{\mathrm{d}\mathbb{P}}=\exp(-\int_0^t rdW_h -\frac{1}{2}\int_0^tr^2 dh)$

My confusion arises from the fact that one could also write: $\mathbb{P}(B):=\mathbb{P}(X(t)\leq b)=\mathbb{P}(W(t)-r\leq b)=\mathbb{P}(W(t)\leq b+r)=\int^{b-r}_{-\infty}f_{W_t}(h)dh$.

So that makes me think that we can discuss probabilistic events related to $X(t)$ and $W(t)$ under the same measure $\mathbb{P}$. However the fact that we can define a measure via a process-specific density then makes me think that it doesn't make sense to have two different processes with different densities under the same probability measure.


1 Answer 1


There is a big confusion about notation and naming in your post....

For clarification let's start by define the probability space $(\Omega,\mathcal{F},\mathbb{S})$ where all of your random variables live on.

Then you assume that $W$ induces a probability measure $\mathbb{P}$ which means nothing else that $$\mathbb{P}(A) := \mathbb{S}(W \in A)$$

This definition is always well defined and now, if there exists a measurable function $f_W \ge 0$ s.t. $$\mathbb{P}(A) = \int_A f_W(h) dh$$ then we call $f_W$ the densitiy of $W$.

Consider that your "definition" $$\mathbb{P}(A):=\int^{a}_{-\infty}f_{W}(h)dh$$ only is valid for sets $A = (-\infty,a)$ and not in general. Luckily on $\mathcal{B}(\mathbb{R})$ is enough to show/have this identity for all $a \in \mathbb{R}$ to get it for all $A \in \mathcal{B}(\mathbb{R})$ (why?) and then we can shortly write $\mathbb{S}(W \le a)$ for $\mathbb{S}(W \in A)$.

Additionally we have then $$f_W(h) = \frac{d}{da} \mathbb{S}(W \le a)$$

Now you consider a second random variable defined by $$X = W - c$$ and a related density $f_X$. So we get a new measure $$\mathbb{Q}(A) := \mathbb{S}(X \in A) = \int_A f_X(h) dh$$ Then again for all sets of the form $A = (-\infty,a)$ we have $$\mathbb{Q}(A) = \mathbb{S}(X \le a) = \int_{-\infty}^a f_X(h) dh$$

And is holds: $$\begin{align} \int_{-\infty}^a f_X(h) dh &= \mathbb{Q}((-\infty,a)) &&= \mathbb{S}(X \le a)\\ &= \mathbb{S}(W-c \le a) &&= \mathbb{S}(W \le a+c) \\ &= \mathbb{P}((-\infty,a+c)) &&= \int_{-\infty}^{a+c} f_W(h) dh \end{align} $$

Additionally we know that the density functions above are nothing else then the R-N-density related to the Lebesgue measure $\lambda$, so we get:

$$\frac{d\mathbb{Q}}{d\mathbb{P}} = \frac{\frac{d\mathbb{Q}}{d\lambda}}{\frac{d\mathbb{P}}{d\lambda}} = \frac{f_W}{f_X}$$

All the arguments above hold in general so if you deal with certain special densities just plug in and calculate…

  • $\begingroup$ Thank you so much for that detailed answer, I really appreciate it. I wanted to clarify a few points if I may please. (i) Looking at $(\Omega,\mathcal{F},\mathbb{S})$, it makes sense to first define $\Omega$: I assume this should be the real line? (ii) We then move onto $\mathcal{F}$: This is the set of all subsets of $\Omega$, and intuitively, all events of interest. I assume these are the Borel sets$ \mathcal{B}(\mathbb{R})$. (iii) What exactly is then $\mathbb{S}$ ? (iv) At what point come the Random Variables into picture? Don't these come before talking about $\mathbb{S}$? $\endgroup$ Jan 20, 2020 at 18:36
  • $\begingroup$ Unfortunately you are wrong… Although we technically work with the underlying probability space $(\Omega, \mathcal{F},\mathbb{S})$ it's not necessary (or sometimes not possible) to know it exactly. When you deal with random variable you are usually not really interested in the probabilty space but the image space. Additionally consider that $\mathbb{Q}$ and $\mathbb{P}$ live on the image space and not on the domain of your random variables. So that's why $\Omega$ in general is not the real line. An additional comment: $\mathcal{B}(\mathbb{R})$ is NOT the set of all subsets! We cannot define… $\endgroup$
    – Gono
    Jan 21, 2020 at 7:21
  • $\begingroup$ a probability measure on the set of all subsets. For the same reason you don't know exactly how $\mathbb{S}$ look like because in standard applications you are only interested (and those are only given) in $\mathbb{P}$ and $\mathbb{Q}$ So you are right: The random variables with their given distributions can come in without talking about $\mathbb{S}$. But you should be aware that technically $\mathbb{S}$ is necessary for a consistent definition as mentioned above. The distributions of your random variables are induced by the r.v. and the (not known) measure $\mathbb{S}$. $\endgroup$
    – Gono
    Jan 21, 2020 at 7:26

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