# Tag Info

8

Scaling volatility as you do is often leading to inaccurate results which is over-estimating volatility especially when you scale daily volatility to even longer periods. Please see the following for more: http://economics.sas.upenn.edu/~fdiebold/papers/paper18/dsi.pdf The above paper also explains why scaling the way you did does not properly account for ...

6

Don't subtract dividends; add them. Add-back the dividends as if they had not been paid out. That will ensure that you have a positive price when deriving the returns. For example, MSFT paid a 0.16 dividend on 2011-15-02. Here are the raw prices, according to Yahoo: date close ---------------- 2011.02.01 27.99 2011.02.02 27.94 2011.02.03 27.65 2011....

3

I'm guessing ${W_t^r}$ and ${W_t}$ correspond to real and nominal endowment at time $t$, respectively, and that ${P_t^g}$ is the price level at time $t$. In that case, $W_t^r \equiv W_t/P_t^g$ follows, and if endowment grows at a nominal interest rate $R_t$, then $W_t = W_{t-1}(1+R_t)$. We can write $W_{t-1}=(W_{t-1}^rP_{t-1}^g)$, so by substitution $W_t=(W_{... 3 When volatility is high, daily volume is high. And when volatility is high, daily returns are high. That's why when volume is high, the price returns are high. Volatility (like volumes) is autocorrelated. This is the phenomenon of volatility clustering (large changes tend to be followed by large changes, of either sign) and volume clustering (large volumes ... 3 Not sure I understand your question. If I have a fixed stream of payments it has some value$V_{fixed}$I can always solve for a spread to LIBOR by simply adding the spread$S$to my calculated stream of LIBOR. That is the value of the LIBOR + spread leg is $$V_{LIBOR}(S) = \sum_{n=1}^{N} D(t_{n}) \alpha(t_{n-1},t_{n}) [L(t_{n-1},t_{n}) + S]$$ where$D(...

3

I think the answer is in your question. Yahoo uses a percentage adjustment for adjusted close prices. So this is the procedure I would do. 1) Calculate the proper return for each day (taking into account splits and divs). 2) Apply the returns going backwards from the current price. By doing it this way it is impossible to get a negative adjusted price, ...

3

The problem is, you're calculating the "return on investment" or "return on original investment". Anytime you are given back ALL or MORE THAN your original investment, you are no longer "invested". As a result, calculating "return on investment" no longer applies. For example with a 100% return of capital, the "original investment" drops to zero. So, ...

3

It looks like 1 and 2 are different portfolios of companies. 1 is a portfolio of dual-listed companies, and 2 is a portfolio of everything in the "market". Once you have constructed these these portfolios, let's say you put the returns for every time step into a vector, call it r, then the average return would be mean(r). You need some clarification as ...

2

The basic CAPM - which is what your regression estimates - says $$R_S = R_f + \beta_S (R_{Market}-R_f)$$ where $$\beta_S = \frac{Cov(R_M,R_S)}{Var(R_M)}$$ i.e. the return of a certain stock depends only on the correlation with the market portfolio. For your pricing equation to work, you will need to have an idea about the expected market (excess) ...

2

The technical analysis point of view: an increase in volume (assuming the price has been in a downtrend) means the crowd are throwing in the towel, i.e. everyone is dumping the stock and assuming that hoped-for rise is now never going to happen. The same on the way up: everyone jumps on the bandwagon. In other words, high volume typically means crowd ...

2

No, it would be $$(RI_{t}-RI_{t-1})/RI_{t-1}$$

2

Actually prices dont make sense as they are correlated with previous samples (prices), returns are not. Better will be difference between prices, but then you dont have reference point and comparability between assets, so eventually you need returns. At the end that is what you are interested in I think as profit is usually measured in return.

2

I think the only valid answer is you can't. The techniques you describe would work of the signal was much stronger than the noise but it seems that with your fund returns this is not the case. You could try to get more data or look at other risk measures like max drawdown to get some idea of the risks involved.

1

To get one-month rate X from three-month rate Y, you use this formula: 1 + X = (1 + Y)^(1/3) To get one-month rate X from annual (12-month) rate Y, you use this formula: 1 + X = (1 + Y)^(1/12) To get three-month rate X from annual (12 month) rate Y, you use: 1 + X = (1 + Y)^(3/12) 0.25% is annual 3% converted to a monthly rate, i.e. (1+0.03)^(1/12) - 1 ...

1

I think there are 2 approaches being a bit mixed up here. You can analyze the option market by looking at implied volatilities and apply Black-Scholes (BS), thus assuming that log-returns follow a Gaussian distribution. Implied volatilies are the parameters that bring together BS and market prices. Then you will observe a pattern of implied volatilies for ...

1

I'm currently also using daily returns which I want to annualize. This is my approach: For every month, I calculate the simple return using the formula: (end-of-month closing price / beginning-of-month closing price) - 1. I use the Excel formula somproduct(geomean(A1:A12+1)-1) to find the monthly compounded return. Finally, I annualize the result of step 2 ...

1

for the square-root rule: it holds for log-returns, if you assume the same variance and no autocorrelation. Because then: $$Var[r_1 + \cdots + r_d] = Var[r_1] + \cdots + Var[r_d] = d Var[r_1]$$ and thus $$\sqrt{Var[r_1 + \cdots + r_d] } = \sqrt{d} \sqrt{Var[r_1]}.$$ This is mathematically true for any distribution that fulfills the assumptions. For the ...

1

It depends on your investment strategy. The most common approach is to use the close price of $p_t$ and $p_{t+1}$. The volatility you measure using this method implies the "assumption" that your are able to trade at close every day. If you choose to compute the daily returns from open to close, then you assume that you are selling your position every night ...

1

No, the "low-beta" anomaly is not the result of the difference between arithmetic and geometric mean returns. Statistical tests verifying the existence of the anomaly rely on models employing the arithmetic mean returns, $$\mu_a = \mu_g + \frac{\sigma^2}{2}$$, hence the penalty excess volatility incurs when compounding returns over time does not explain the ...

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