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| bio | website | researchgate.net/profile/… |
|---|---|---|
| location | Vienna | |
| age | 31 | |
| visits | member for | 11 months |
| seen | May 14 at 7:53 | |
| stats | profile views | 130 |
Risk Manager at Raiffeisen Capital Management
External Lecturer at Vienna University of Technology
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Mar 15 |
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Is Optimization ignoring correlation valid? @Richard I have found a link to forex. They calculate various correlations between currencies. I don't know all details of their calculations but it seems that the correlations are highly unstable. This would mean that projecting them to the future causes a bias. Could this be true? Can you tell us something (graph, numbers) about the correlations of USD/EUR and USD/JPY e.g. are these numbers very instable? Are the volatilies more stable? |
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Mar 15 |
revised |
Is Optimization ignoring correlation valid? added 160 characters in body |
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Mar 14 |
awarded | Nice Question |
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Mar 14 |
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Correlation decay in lognormal distribution Ok, now I understand. The two prices have correlated log-returns and the correlation of the prices themselves decays. Interesting ... I have to check. |
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Mar 14 |
answered | Is Optimization ignoring correlation valid? |
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Mar 14 |
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Correlation decay in lognormal distribution Would you mind writing down some formulas? You have 2 correlated geometric BM. Then you look at the correlation of the returns, $ r_i = \mu_i dt + \sigma_i dB_t^i$ for $i =1,2$ with $B^1$ and $B^2$ correlated, I guess. I have to check myself but I don't see why this correlation should vanish. Am I missing an important point? |
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Mar 13 |
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Calculate the expectation of a shift CDF In the question it starts with $X \sim N(0,\sigma^2)$ and we want to know the expectation of $F_{X_1}(X_1 + a)$ for $a>0$ @nkhuyu is this correct? If yes, then $X_1$ and $X_2 = X_1 + a$ are far from independent. |
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Mar 13 |
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Calculate the expectation of a shift CDF @AlexeyKalmykov Yes ... but this is important. $X_1$ and $X_2$ can have the same law. but $X_1$ and $X_2 = X_1 +a$ are simply one and the same if $a = 0$. |
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Mar 13 |
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Calculate the expectation of a shift CDF Shouldn't you analyse $P[F_{X_1}(X_2) \le x]$? and $F_{X_1}$ is a deterministic distribution function. |
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Mar 13 |
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Calculate the expectation of a shift CDF If $X_2 = X_1 + a$ don't you get $F_{X_1}(X_2)=P[X_1 \le X_1 + a] = 1$? |
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Mar 13 |
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Calculate the expectation of a shift CDF Hi, I like your approach, but there are problems, I think. Take the unshifted case. $P[X_1 - X_2 \le 0]$ .. but in the unshifted case $X_1 = X_2$ and therefore $X_1-X_2 = 0$. So this is true if $X_1$ and $X_2$ are different Gaussian random variables with mean $0$ and the same variance. |
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Mar 13 |
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Calculate the expectation of a shift CDF @GoodGuyMike Sorry to say, but this looks too complicated. Alexey's approach in the other answer is more direct (I tried something similar). |
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Mar 13 |
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Calculate the expectation of a shift CDF All that I have tried ended up here too. Yes, maybe there is no (at least no easy) closed-form solution. The MC simulation persuades me ... |
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Mar 12 |
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Calculate the expectation of a shift CDF I tried to do the calculation for the specific case ($\mu=0$) and wanted to reduce it to the unshifted one ... but I did not manage. |
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Mar 12 |
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Calculate the expectation of a shift CDF @John, it definately should ... I just didn't have time for the proof. The answer must be very similar to the unshifted case. |
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Mar 12 |
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Calculate the expectation of a shift CDF I think the question fits for both as it could really be asked in a quant interview. In my mind it is ok. |
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Mar 12 |
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Calculate the expectation of a shift CDF Wait a moment. Isn't it true that $F(X)$ is uniform if $X$ has a continuous density. Therefore $E[F(X)] = 1/2$? I am not sure whether this helps us for $E[F(X+a)]$ ... |
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Mar 12 |
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Calculate the expectation of a shift CDF You say that you know how to calculate $E[F(X)]$, where $F$ is the distribution function of $F$, right? What is the trick. If you show us this, then we can work on $E[F(X+a)]$. |
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Mar 11 |
answered | Square root of time |
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Mar 11 |
revised |
Predict Market Direction, What is forecastable/unforecastable? Changes $t+1$ to $t$. |
