0

Let's suppose that $X_t$ and $Y_t$ are random variables, which are $\mathcal{F}_t$-measurable, $\mathcal{F}_t$ is a filtration. Suppose also that random variables $Z$ and $Y_t$ are independent. Does it mean that also $Z$ and $X_t$ are independent? If yes, how should I prove it? I'm considering standard probability space. Thank you in advance for your help

siwy9
  • 125

1 Answers1

1

No, basically because the mere fact that $X$ and $Y$ are both measurable with respect to a common sigma algebra tells us very little about $X$ and $Y$. For example, if the underlying sigma algebra is the power set, then all variables are measurable, and of course we can find three variables $X$, $Y$ and $Z$ such that $Z$ is independent to $Y$ but not $X$.

Your proposition is true if the underlying sigma algebra is, for example, the sigma algebra generated by $Y$, so that $X$ is $Y$-measurable and therefore more or less (but not really) a function of $Y$. Obviously if $Z$ is independent to $Y$ it's going to be independent to a function of $Y$.

Jack M
  • 27,819
  • 7
  • 63
  • 129
  • Thank you, that makes sense. I have one additional question. Does assumption that $Y_t$ is a martingale with respect to $\mathcal{F}_t$ change anything? – siwy9 Feb 14 '18 at 21:15
  • @siwy9 No. Look more closely at the question you're asking. You're not even using the fact that $F_t$ is a filtration, in fact, $t$ is fixed for the whole question, so you're not even using the fact that you have a whole family of sigma algebras rather than just one. As far as your question is concerned, you may as well write $X$ and $Y$ instead of $X_t$ and $Y_t$, and replace the sigma algebra $F_t$ with an arbitrary sigma algebra $F$. – Jack M Feb 15 '18 at 13:14