Expectation of a monotone function of CDF: $\mathbb E \left [g(F(X)) \right ]$

Question

If $X$ is a random variable with distribution function $F,$ then $\mathbb{E}[(F[X])^{-1/2}]$ can be computed by integration by parts, if $X$ has a continuous density $f$. What happens in the general case? Can we always compute it (or bound it)?

Answer 1

From here, we know that when $F(x)$ is the continuouse CDF of $X$, $$F(X)\sim U(0,1).$$

Hence, you can compute the expectation of any measurable function $g$ of $F(X)$ (if it is finite) as follows:

$$\mathbb E \left [g(F(X)) \right ]=\int_{0}^1 g(x)\text{d}x.$$

For $$g(x)=x^{-a}$$ with $a \ge 1$, it is infinite. For $0<a<1$, it becomes

$$\frac{1}{1-a}.$$

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Responding to @Beleth, if $F(x)$ is not continuouse, unfortunatly the above dose not hold. The most similar result holds for the following transform: $G:\mathbb{R}\times[0,1]\rightarrow\mathbb{R}$ of $X$:

$$G(x,v)=F(x-)+v(F(x)-F(x-)).$$

For $V\sim U(0,1),$ which is independent of $X$, we have

$$G(X,V)\sim U(0,1),$$

which reduces to $F(X)\sim U(0,1) $ for continuous CDF $F$.

Update:

From the above one can see that:

$$\color{blue}{F(X)=G(X,V)+(1-V)(F(X)-F(X^-)) \ge_{1st} G(X,V) \sim U(0,1)},$$

where $\ge_{1st}$ denotes the first order stochastic dominance. Thus, for an increasing function $g$ we have:

$$ \fbox{$\int_{0}^1 g(x)\text{d}x \le \mathbb E \left [g(F(X)) \right ] \le g(1)$},$$

and for a decreasing function $g$:

$$\fbox { $g(1) \le \mathbb E \left [g(F(X)) \right ] \le \int_{0}^1 g(x)\text{d}x. $ } $$

The upper bound in the first and the lower bound in the second are attained when $F(x)$ is a discrete CDF with a single jump, and the lower bound in the first and the upper bound in the second are attained when $F(x)$ is a continuous CDF, as shown earlier above.

In particular, for $g(x)=x^{-a}$ with $0<a<1$, we have

$$1 \le \mathbb E \left [g(F(X)) \right ] \le \frac{1}{1-a}.$$

Answer 2

Let $Y = \frac{1}{\sqrt{F(X)}}$. Since $Y$ is non-negative, we have $$ \mathbb{E}[Y] = \int_{0}^\infty \mathbb{P}\{ Y \geq x \} dx = \int_0^1 1\cdot dx + \int_{1}^\infty \mathbb{P}\{ Y \geq x \} dx = 1+ \int_{1}^\infty \mathbb{P}\{ Y \geq x \} dx \tag{1} $$ the second-to-last equality since $Y \geq 1$ almost surely. Now, this gives us: $$\begin{align*} \mathbb{E}[Y] &= 1+ \int_{1}^\infty \mathbb{P}\left\{ \frac{1}{\sqrt{F(X)}} \geq x \right\} dx\\ &= 1+\int_{1}^\infty \mathbb{P}\left\{ F(X) \leq \frac{1}{x^2} \right\} dx \end{align*}$$ Now, if $F$ was a continuous CDF, then as in Amir's answer we could say that $F(X)$ is uniform on $[0,1]$, and then $\mathbb{P}\left\{ F(X) \leq {1}/{x^2} \right\} = 1/x^2$; so that the integral would be 1, and the overall expression $1+1=2$.

This is however not the case in general. Yet, we do have the inequality $$ \mathbb{P}\left\{ F(X) \leq {1}/{x^2} \right\} \leq 1/x^2 $$ (See this question), so that $$\begin{align*} \mathbb{E}[Y] &\leq 1+\int_{1}^\infty \frac{dx}{x^2} =2. \end{align*}$$

Conclusion: $$ \boxed{1 \leq \mathbb{E} \frac{1}{\sqrt{F(X)}} \leq 2} $$ for every r.v. $X$.

Moreover, this is tight: the lower bound is achieved for a point mass, and the upper bound for any r.v. with strictly increasing continuous cdf.