Wie berechnet man die Erwartung von

Wenn $X_i$ exponentiell verteilt ist $(i=1,...,n)$ mit dem Parameter $\lambda$ und $X_i$ ‚s ist voneinander unabhängig, was die Erwartung

$$\left(\sum_{i=1}^n {X_i} \right)^2$$

in Bezug auf $n$ und$\lambda$ und möglicherweise andere Konstanten?

Hinweis: Diese Frage wurde unter /math//q/12068/4051 mathematisch beantwortet . Die Leser würden es sich auch ansehen.

Wenn , dann (unter Unabhängigkeit), y = Σ x i ~ G a m m a ( n , 1 / λ ) , so dass y ist gamma verteilt (siehe wikipedia ). Also brauchen wir nur E [ y 2 ] . Da V a r [ y ] = E [ y 2 ] - $x_i \sim Exp(\lambda)$$y = \sum x_i \sim Gamma(n, 1/\lambda)$$y$$E[y^2]$ , wissen wir, dass E [ y 2 ] = V a r [ Y ] + E [ y ] 2 . Daher ist E [ y 2 ] = n / λ 2 + n 2 / λ 2 = n ( 1 + n ) / λ 2 (sieheWikipediafür die Erwartung und Varianz der Gammaverteilung).$Var[y] = E[y^2] - E[y]^2$$E[y^2] = Var[y] + E[y]^2$$E[y^2] = n/\lambda^2 + n^2/\lambda^2 = n(1+n)/\lambda^2$

Die obige Antwort ist sehr schön und beantwortet die Frage vollständig, aber ich werde stattdessen eine allgemeine Formel für das erwartete Quadrat einer Summe bereitstellen und sie auf das hier erwähnte spezifische Beispiel anwenden.

Für jeden Satz von Konstanten ist es eine Tatsachedass$a_1, ..., a_n$

$$\left( \sum_{i=1}^{n} a_i \right)^2 = \sum_{i=1}^{n} \sum_{j=1}^{n} a_{i} a_{j}$$

this is true by the Distributive property and becomes clear when you consider what you're doing when you calculate $(a_1 + ... + a_n) \cdot (a_1 + ... + a_n)$ by hand.

Therefore, for a sample of random variables $X_1, ..., X_n$, regardless of the distributions,

$$E \left( \left[ \sum_{i=1}^{n} X_i \right]^2 \right) = E \left( \sum_{i=1}^{n} \sum_{j=1}^{n} X_i X_j \right) = \sum_{i=1}^{n} \sum_{j=1}^{n} E(X_i X_j)$$

provided that these expectations exist.

In the example from the problem, $X_1, ..., X_n$ are iid ${\rm exponential}(\lambda)$ random variables, which tells us that $E(X_{i}) = 1/\lambda$ and ${\rm var}(X_i) = 1/\lambda^2$ for each $i$. By independence, for $i \neq j$, we have

$$E(X_i X_j) = E(X_i) \cdot E(X_j) = \frac{1}{\lambda^2}$$

There are $n^2 - n$ of these terms in the sum. When $i = j$, we have

$$E(X_i X_j) = E(X_{i}^{2}) = {\rm var}(X_{i}) + E(X_{i})^2 = \frac{2}{\lambda^2}$$

and there are $n$ of these term in the sum. Therefore, using the formula above,

$$E \left( \sum_{i=1}^{n} X_i \right)^2 = \sum_{i=1}^{n} \sum_{j=1}^{n} E(X_i X_j) = (n^2 - n)\cdot\frac{1}{\lambda^2} + n \cdot \frac{2}{\lambda^2} = \frac{n^2 + n}{\lambda^2}$$

is your answer.

This problem is just a special case of the much more general problem of 'moments of moments' which are usually defined in terms of power sum notation. In particular, in power sum notation:

$$s_1 = \sum_{i=1}^{n} X_i$$

Then, irrespective of the distribution, the original poster seeks $E[s_1^2]$ (provided the moments exist). Since the expectations operator is just the 1st Raw Moment, the solution is given in the mathStatica software by:

[ The '___ToRaw' means that we want the solution presented in terms of raw moments of the population (rather than say central moments or cumulants). ]

Finally, if $X$ ~ Exponential($\lambda$) with pdf $f(x)$:

f = Exp[-x/λ]/λ;      domain[f] = {x, 0, ∞} && {λ > 0};

then we can replace the moments $\mu_i$ in the general solution sol with the actual values for an Exponential random variable, like so:

All done.

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P.S. The reason the other solutions posted here yield an answer with $\lambda^2$ in the denominator rather than the numerator is, of course, because they are using a different parameterisation of the Exponential distribution. Since the OP didn't state which version he was using, I decided to use the standard distribution theory textbook definition Johnson Kotz et al … just to balance things out :)