Proof of $E[Y]=\int_0^x nX^{n-1}(1-F_X(x))dx$ for $Y=X^n$ [duplicate]

Solution 1:

For your specific question, notice that

$$ 0 \leq x^r (1 - \mathbb{P}(X \leq x)) = x^r \mathbb{P}(X \geq x) = x^r \int_{x}^{\infty} f(s) \mathrm{d}s \leq \int_{x}^{\infty} s^r f(s) \mathrm{d}s. $$

Since we know that $\mathbb{E}[X^r] = \int_{0}^{\infty} s^r f(s) \, \mathrm{d}s$ is finite, it follows that $\int_{x}^{\infty} s^r f(s) \mathrm{d}s \to 0$ as $x\to\infty$. So, the quantity in question vanishes as $x\to\infty$.

For a more systematic approach, we can utilize Tonelli's theorem. (This is exactly as in @Marios Gretsas's answer, but let met recap the argument for the sake of self-containedness.)

Tonelli's theorem asserts the following: For the iterated integral of a non-negative function, we can always interchange the order of integration regardless of its finiteness. So, for any non-negative random variable $X$ and for any $r > 0$,

\begin{align*} \mathbb{E}[X^r] = \mathbb{E}\bigg[ \int_{0}^{X} rx^{r-1} \, \mathrm{d}x \bigg] &= \mathbb{E}\bigg[ \int_{0}^{\infty} rx^{r-1} \mathbf{1}( x < X ) \, \mathrm{d}x \bigg] \\ &= \int_{0}^{\infty} rx^{r-1} \mathbb{E}\big[ \mathbf{1}( x < X ) \big] \, \mathrm{d}x \tag{Tonelli}\\ &= \int_{0}^{\infty} rx^{r-1} \mathbb{P}(X > x) \, \mathrm{d}x. \end{align*}

Neither density assumption nor finiteness of $r$-th moment is used here.

Solution 2:

As another proof you can do this.

$$\int_{0}^{\infty} rx^{r-1}P(X>x) dx$$ $$=\int_{0}^{\infty} \int1_{\{X>x\}}(y)rx^{r-1}dP(y)dx$$ $$=^{TONELLI }\int \int_{0}^{ X(y)} rx^{r-1}dxdP(y) =\int X^r(y)dP(y)$$

Solution 3:

You need to be careful here not to use $\infty-\infty$ gymnastics. The survival function $S(x):=\Bbb P(X>x)$ satisfies $S^\prime=-f(x)$, so $$\Bbb E(X^r)-\int_0^\infty rx^{r-1}S(x)dx=-\int_0^\infty(x^rS^\prime+(x^r)^\prime S)dx=[-x^rS]_0^\infty=-\lim_{x\to\infty}x^rS(x).$$That just leaves proving this limit is $0$.