probability - Show the following relation: $mathbb{E}[X^{q}mathrm{1}_{{X>a}}]=a^{q}P(X>a)+qint_{a}^{infty}P(X>x)x^{q-1}dx $

Let $X$ be a non negative random variable, $a>0$ be a constant and $q\ge 1$ be a positive integer. How can I show the following: \begin{equation} \mathbb{E}[X^{q}\mathrm{1}_{\{X>a\}}]=a^{q}P(X>a)+q\int_{a}^{\infty}P(X>x)x^{q-1}dx ? \end{equation}

I have tried the computation using the complementary CDF trick

$$\mathbb{E}[X^{q}\mathrm{1}_{\{X>a\}}]=\int_{a}^{\infty}P(X^{q}>t)dt=q\int_{a^{1/q}}^{\infty}P(X>u)u^{q-1}du$$

which doesn't help. Any ideas appreciated.

Answer

You made two mistakes, both of which seem to be from trying to move too quickly.

First, we should have $$\mathbb{E}[X^q 1_{\{X>a\}}] = \int_0^\infty P(X^q 1_{\{X>a\}} > t) \,dt$$ and we should split the integral into two integrals, but the minimum $t$-value at which $$\{X^q 1_{\{X>a\}} > t\} = \{X^q > t\}$$ is not $t=a,$ it's actually $t=a^q.$

Therefore, the above equality becomes $$\mathbb{E}[X^q 1_{\{X>a\}}] = \int_0^{a^q} P(X^q1_{\{X>a\}} > t)\,dt + \int_{a^q}^\infty P(X^q > t)\,dt$$

Now simplify each of these. For the first, simplify the integrand and for the second, use substitution.