Show the expected value of a function is greater than the expected value of another

Question

Consider two distributions

$$F(x) =\begin{cases} 0 &\text{if} \ \ x < 0\\ \frac{1}{4} &\text{if} \ \ 0\leq x < 2\\ 1 &\text{if} \ \ 2\leq x \end{cases}$$ and $$G(x) =\begin{cases} 0 &\text{if} \ \ x < 1\\ 1 &\text{if} \ \ 1\leq x \\ \end{cases}$$ Show that the expected value of $x$ under $F$ is greater than the expected value of $x$ under $G$ but that $F$ does not first-order stochastically dominate $G$.

Attemped solution - It seems the expected value of both distributions will be infinity, so I do not know how to show the latter. For first-order stochastic dominance we have $$U(F) = \int_{-\infty}^{0}dx + \int_{0}^{2}\frac{1}{4}dx + \int_{2}^{\infty}dx$$ again it does not seem that we will get any meaningful evaluation.

Answer 1

I'll give you the details that you need for this and have you fill in any details (or ask me, and I can put them in). Let $\mathbb{E}_{F}[X]$ be the expected value of $X$ under $F$, and similarly for $\mathbb{E}_{G}[X]$.

Notice that $F$ is a distribution function and not a probability mass/density function - so you would not be integrating or summing $F$, but integrating or summing the probability density or probability mass function, respectively. This applies similarly to $G$.

$F$ is constant throughout, so that means there are point masses. The probability mass function corresponding to $F$ consists of a probability mass of $\dfrac{1}{4}$ at $x = 0$ and a probability mass of $\dfrac{3}{4}$ at $x = 2$. Hence, $$\mathbb{E}_{F}[X] = \dfrac{1}{4}(0)+\dfrac{3}{4}(2)=\dfrac{6}{4}=1.5\text{.}$$

Similarly for $G$, there is a probability mass of $1$ at $x = 1$, hence $$\mathbb{E}_{G}[X] = 1(1) = 1\text{.}$$ Hence, $\mathbb{E}_{F}[X] > \mathbb{E}_{G}[X]$.

$F$ does not first-order stochastically dominate $G$. If this were the case, then $F(x) \leq G(x)$ for all $x$ with strict inequality at some $x$. However, for $x \in [0, 1)$, $F(x) = \dfrac{1}{4}$ and $G(x) = 0$ so $F(x) > G(x)$.