Timeline for How can I use the Radon-Nikodym theorem to show that forward measure is indeed measure?
Current License: CC BY-SA 4.0
16 events
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| Apr 6, 2020 at 19:13 | comment | added | LazyCat | I certainly didn't mean it to become a pissing contest. I was surprised by your rather involved answer to what seemed to be such a simple question, and assumed I am missing something obvious. Doesn't seem to be the case so far. Yes, you are reading misreading my comments. In the first comment, I meant the function $f$ in the OP's question which is already normalized, in the later one I meant a general function, which still needs to be normalized. | |
| Apr 6, 2020 at 19:06 | comment | added | Daneel Olivaw | And anyway, I really don't see what your point is: why don't you simply put this into an answer? | |
| Apr 6, 2020 at 19:04 | comment | added | Daneel Olivaw | Terrible mistake, I was thinking about the expectation. But you can simply take $f(x)=|x|$. Don't you see your first comment is wrong? Unless I am totally misinterpreting your notation, what you are saying is: "the expectation of any non-negative measurable function $f$ under a probability measure is always equal to $1$" (that is literally what is written in your comment). If you are changing measures, be specific, otherwise it's ambiguous. In fact, you've amended yourself. Which one is it, $\nu(A)=1/Z\int_A f$ or $\nu(A)=\int_A f$? Which measures? etc. | |
| Apr 6, 2020 at 18:56 | comment | added | LazyCat | I appreciate that you try to make the discussion constructive, but the integral of Gaussian density over R is definitely one, not zero. | |
| Apr 6, 2020 at 18:31 | comment | added | LazyCat | what is not true? That $\nu(A) = \frac1Z \int_A f,$ where $Z = \int_\Omega f$ is a probability measure? | |
| Apr 6, 2020 at 18:24 | history | edited | Daneel Olivaw | CC BY-SA 4.0 |
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| Apr 6, 2020 at 17:04 | comment | added | Daneel Olivaw | @LazyCat regarding your comment, then it’s misleading: that is generally not true, it is true for $f$ constructed as a ratio of numéraires. And it is true because of the martingale property under the risk-neutral measure. | |
| Apr 6, 2020 at 16:57 | comment | added | Daneel Olivaw | This discussion is extending unnecessarily and I am not sure what the endpoint is. Yes, the denominator is chosen so that the RN derivative’s expectation is 1: is the OP aware that this is a condition for the RN derivative to be well-defined? I don’t know, I gave an answer quite detailed so that the construction of new measures with other numéraires is sufficiently clear. Please feel free to give an alternative answer. The more derivations posted, the richer the resources of this site will be. | |
| Apr 6, 2020 at 16:31 | comment | added | LazyCat | Well, my point is that unless I am mistaken your response is much more complicated than needs be. For your second question - this is precisely how the denominator is in OP's formula is chosen - to normalize the whole thing to 1. | |
| Apr 6, 2020 at 16:06 | comment | added | Daneel Olivaw | @LazyCat your obvious observation is actually missing a critical assumption: why is it obvious that $\int_\Omega fd\mu=1$? | |
| Apr 6, 2020 at 15:59 | comment | added | Daneel Olivaw | Well, OP asks the question so he obviously does not find it trivial or wants further details; and there are theorems out there, such as Radon-Nikodym, carefully establishing these results and under what conditions they hold. | |
| Apr 6, 2020 at 15:37 | comment | added | LazyCat | which part in my comment is not obvious? I feel like it's a trivial and very generic observation. | |
| Apr 6, 2020 at 15:34 | comment | added | Daneel Olivaw | In particular, the OP asked about how to verify $Q_T$ is a well-defined probability measure using the RN theorem, but the RN theorem is unapplicable in that context. | |
| Apr 6, 2020 at 15:29 | comment | added | Daneel Olivaw | @LazyCat what you say is correct, but it is not straightforward: it is actually a theorem. What I am doing is displaying some steps on how the underlying argument goes. I am also showing why the ratio of numéraires is a well-defined Radon-Nikodym derivative. I am also making clear the construction of the RN derivative along t. | |
| Apr 6, 2020 at 15:18 | comment | added | LazyCat | Maybe I am missing something, but I feel like this way overcomplicates the answer. If you have a measure $\mu$ on some measurable space $\Omega$ and a non-negative measurable function $f,$ then $\nu(A) = \int_A f$ will define always measure. In this case, it's also obvious, that $\int_\Omega f = 1,$ so it's a probability measure. | |
| Apr 6, 2020 at 14:13 | history | answered | Daneel Olivaw | CC BY-SA 4.0 |