Abweichung der Jahresrendite basierend auf der Abweichung der monatlichen Rendite

11

Ich versuche, die ganze Varianz / Standardfehler-Sache einer Zeitreihe von Finanzrenditen zu verstehen, und ich denke, ich stecke fest. Ich habe eine Reihe von monatlichen Aktienrenditedaten (nennen wir es ), die einen erwarteten Wert von 1,00795 und eine Varianz von 0,000228 haben (Standardabweichung ist 0,01512). Ich versuche, den schlechtesten Fall der jährlichen Rendite zu berechnen (sagen wir den erwarteten Wert minus dem doppelten Standardfehler). Welcher Weg ist der beste Weg, dies zu tun? A . Berechnen Sie es für einen einzelnen Monat ( μ X - 2 σ X = 0,977 ) und multiplizieren Sie es 12 Mal mit sich selbst (= 0,7630 ). B . Angenommen, die Monate sind unabhängig, definieren Sie Y = X

μX2σX=0.977

12-mal, finde den erwarteten Wert E [ Y ] = ( E [ X ] ) 12 ) und die Varianz var [ Y ] = ( var [ X ] + ( E [ X ] ) 2 ) 12 - ( ( E [ X. ] 2 ) 12Y=XX...XE[Y]=(E[X])12var[Y]=(var[X]+(E[X])2)12((E[X]2)12 Der Standard-Dev ist in diesem Fall 0,0572 und der erwartete Wert minus dem doppelten Standard-Dev ist 0,9853 .

C . Multiplizieren Sie den monatlichen Standard. dev mit , um die jährliche zu erhalten. Verwenden Sie es, um den schlechtesten Jahreswert (μ-2σ) zu ermitteln. Es kommt als0,9949 heraus. Was ist richtig? Was ist der richtige Weg, um den erwarteten Jahreswert minus dem doppelten Standard zu berechnen? dev wenn du diese eigenschaften nur für die monatlichen daten kennst? (InRegel - wennY=XX...X12mal undμX,σXsind bekannt, wasμY-2σY?)12μ2σ

Y=XX...XμXσXμY2σY

Lyosef
quelle

Antworten:

7

ΔP/P=(Pt+1Pt)/PtP250250 to annualize them. This corresponds to your case C. The point here is to rescale so that a meaningful annual figure can be reported from daily figures (but you wouldn't use this to rigorously compare metrics derived from daily against those derived from monthly). In general, you'd do all your calculations and make all your decisions at the frequency you collected your data (monthly in your case).

The theoretically correct approach is to use log returns = log(Pt+1/.P.t)(mit natürlichen Protokollen). Die Formel für die Erwartung einer Summe von Zufallsvariablen kann dann korrekt verwendet werden, da die Summe der Protokollrückgaben das Protokoll des Produkts der Rückgaben ist.

Wenn Sie Protokollrückgaben verwenden, gibt der zentrale Grenzwertsatz eine theoretische Rechtfertigung dafür, dass die Protokollrückgaben normal verteilt sind (im Wesentlichen besagt der zentrale Grenzwertsatz, dass die Summe der unabhängigen Variablen zu einer Normalverteilung tendiert, wenn die Anzahl der Zufallsvariablen in der Summe zunimmt ). Auf diese Weise können Sie eine Wahrscheinlichkeit für eine Rendite von weniger als zuweisenμ- -2σ (Die Wahrscheinlichkeit ergibt sich aus der kumulativen Verteilungsfunktion für die Normalverteilung: Φ(- -2)0,023). If the log returns are normally distributed, then we say that the returns are lognormally distributed -- this is one of the assumptions used deriving the famous Black Scholes option pricing formula.

One thing to note is that when a proportional return is small, then the proportional return is approximately equal to log returns. The reason for this is that the Taylor series for the natural logarithm is given by log(1+x)=x12x2+13x3+, and when the proportional return x is small you can ignore terms with x2, x3, etc. This approximation gives a bit more comfort to those who choose to work with proportional returns and multiply the mean by n and the standard deviation by n!

You ought to be able to find further information on the web. E.g., I tried searching for "log returns" to refresh my memory, and the first hit seemed pretty good.

What you have put in case A is wrong. In the rest of your post you use the facts that (i) the expectation of a sum of random variables is the sum of their expectations, and (ii) the variance of a sum of independent random variables is the sum of their variances. From (ii), it follows that the standard deviation of n independent identically distributed random variables with standard deviation σ is nσ. But in case A you have multiplied both the mean μX and the standard deviation σX by n, whereas the mean needs to be multiplied by n and the standard deviation by n.

A subtle but important point, as noted by in @whuber's comment, is that rule (ii) requires correlation, which in the case of time series means no serial correlation (usually true but worth checking). The requirement for independence holds in both the proportional and log returns case.

(I haven't seen case B, the product of random variables, before. I don't think this approach is commonly used. I haven't looked in detail at your calculations, but your numbers look about right, and the formula can be found on wikipedia. In my opinion this approach seems a lot more complicated than either the approximation involved in using proportional returns or the theoretically sound approach of using log returns. And, compared to using log returns, what can you say about the distribution of Y? How can you assign probabilities to your worst case return, for example?)

TooTone
quelle
1
+1 Using logs is the key. It may be worth noting the implicit assumption in both the question and this answer that the monthly returns exhibit no appreciable serial correlation. (In my experience that's a reasonable assumption for most financial time series, but it's always worth checking.)
whuber
Thank you on the log-return suggestion! I will look it up. However - regarding the rest of your answer - in my post I actually calculated P_t+1/P_t (and not [Pt+1-Pt /Pt]), so the expected value 1.00795 actually means return of 0.795%. That's why I multiplied the monthly values and not added them. (So the annual value in A is actually the monthly "worst-case" value to the power of 12). I would be glad to know if now you think differently on A or B, given that my question refers to a product of random variables and not their sum. Again, thanks a lot.
lyosef
1
@NightMaster769 Sorry, I should have referred more directly to your post. I realized you were multiplying in order to compound the returns correctly but I didn't state that explicitly. This is after all why you were rightly concerned about using the formulas for adding random variables. Nevertheless, A just compounds the "2 standard deviation monthly bad return" over 12 months. It doesn't give you "2 standard deviation annual bad return". Regarding B, your approach seems sound but is complicated compared to log returns, and begs the question "What is the distribution of Y?".
TooTone
@whuber Thanks I've added your point about serial correlation.
TooTone