Expected Value of a Generic Positive Random Variable

Here is a math problem:

Suppose we are told that P(X > 0) = 1 and are given that \lim_{n\to\infty} x\big[1-F(x)\big] = 0 . Show that \mathbb{E}[X] = \int_0^{\infty} P(X > x)dx .

Let’s start with the definition of expectation and use integration by parts.

\mathbb{E}[X] = \int_{-\infty}^{\infty}xf_X(x)dx = \int_0^{\infty}xf_X(x)dx since we are given x \in (0,\infty) .

Now using integration by parts, making the natural selection we have:

u=F_X(x) \quad \quad v=x
u'=f_X(x) \quad \quad v'=dx

Recall that \int u'v = uv-\int uv' and plugging in our selections we get:

\mathbb{E}[X] = \int_0^{\infty}xf_X(x)dx = xF_X(x)\bigm|_0^{\infty} - \int_0^{\infty} F_X(x)dx

Let’s rewrite \int_0^{\infty} F_X(x)dx as \int_0^{\infty} \big(1 - P(X > x)\big)dx . This simplifies to \int_0^{\infty} dx - \int_0^{\infty} P(X > x)dx . Let’s plug this back into our equation above:

\mathbb{E}[X] = \int_0^{\infty}xf_X(x)dx = xF_X(x)\bigm|_0^{\infty} - \bigg(\int_0^{\infty} dx - \int_0^{\infty} P(X > x)dx \bigg)

\mathbb{E}[X] = xF_X(x)\bigm|_0^{\infty} - x\bigm|_0^{\infty} + \int_0^{\infty} P(X > x)dx

\mathbb{E}[X] = \big[xF_X(x)- x\big]_0^{\infty} + \int_0^{\infty} P(X > x)dx

\mathbb{E}[X] = \big[x(F_X(x)- 1)\big]_0^{\infty} + \int_0^{\infty} P(X > x)dx

\mathbb{E}[X] = \big[-x(1-F_X(x))\big]^{x=\infty} + \int_0^{\infty} P(X > x)dx since plugging in 0 to the first half of our expression just yields 0 . We can’t actually evaluate x at infinity, instead we take the limit:

\mathbb{E}[X] = -\lim_{x\to\infty} x\big[1-F(x)\big] + \int_0^{\infty} P(X > x)dx , but recall that we are told \lim_{x\to\infty} x\big[1-F(x)\big] = 0 and so we are simply left with:

\mathbb{E}[X] = \int_0^{\infty} P(X > x)dx , our desired result.


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