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One more kind of problem in your basket: $$\max_w \left(w^T \mu -q \cdot w^T \Sigma w\right)$$ where $q\geq 0$ is a risk-aversion parameter. In case $q\to\infty$ you are extremely risk-averse, and you minimize the variance without caring about the mean. If $q =0$ you are risk-neutral, and you're only interested in maximizing the mean. You can put all ...

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As a practitioner, I have worked on the following Maximize Yield/OAS for a Fixed Income Portfolio keeping the Rates Duration (Key Rate Durations) and Spread duration in a constrained range . There are other constraints such as No short selling Max amount you can buy is X% of Max outstanding amount in market Maximum exposure to a perticular country , ...

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Intuitively, I would be thinking about how SP500 and SP1500 are different. A few points that come to my mind: Different liquidity of constituents Number and frequency of rebalancing/constituents changes Different levels of diversification Different cost of matching the index Since, again intuitively (EDIT: and a little less so since I have added few ...

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You are asking two different questions: what would be the model result, and what would be the actual performance of an actual portfolio. The optimal model results with the S&P 1500 will be at least as good as the model results with the S&P 500. The S&P is a proper subset of the S&P 1500, so you can get the results of the S&P 500 model by ...

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Well, you are asking something very subjective. In addition it should be mentioned that S&P500 are the companies with higher capitalization of S&P1500. Therefore a huge weight of S&P1500 is set by S&P500. In fact, as it can be seen in 2008 both went down a 37%, in the other hand S&P500 has 80% of the total of the US equity Market. After ...

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On the same time period or different ones? It's difficult to say for a different time period. It's actually difficult to say for the same time period, because dynamics are non-stationary. Let's think about it like this: say you perform mean-variance on the S&P1500, with a short time period: this implies that your estimate of the covariance matrix is ...

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