# Moving expectation into probability [closed]

Let $F(g), Y$ be a random variables with finite expected values. Then, we define $X(g)$ as:

$$X(g) := \left\{ \begin{array}{ll} 1, & \quad \text{if } F(g) + Y > 0 \\ 0, & \quad \text{else.} \end{array} \right.$$

Is it true that $\mathbb{E} [ X(g) ] = \text{Pr} [\mathbb{E}[F(g)] + \mathbb{E}[Y] > 0]$?

## closed as unclear what you're asking by Xi'an, mdewey, Michael Chernick, John, GreenparkerApr 26 '17 at 9:35

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• That seems to be correct. – Michael Chernick Apr 24 '17 at 2:10
• @Marius I'm not sure I follow your logic. How does that lead to the equality I asked about? – Jess Smith Apr 24 '17 at 2:50

## 2 Answers

The expectation of the random variable is the function of parameters, and is not random variable anymore. So Pr{E(...)>0} does not exist.

• Instead of "does not exist" I would say the probability $P(\textrm{constant}>0)$ exists and equals either $0$ or $1$... (formalized as $P(\{\omega \in \Omega \mid \textrm{constant} > 0\})$, this is either $P(\emptyset)$ or $P(\Omega)$ depending on the constant). – Juho Kokkala Apr 24 '17 at 16:56

Using the law of total probability, $$\mathbb{E} [ X(g) ] = \text{P}(\mathbb{F(g) + Y > 0})$$ There should not be any expectations inside the probability.