Expected Value

Suppose that

X

is a discrete random variable then we define

E (X) : = \sum x P (X = x)

Squared Expection

Suppose

X

is a random variable such that

P (X = - 2) = 0.2, P (X = 1) = 0.5, P (X = 2) = 0.3

and

P (X = a) = 0

for all other values. Compute

E (X^{2})

and show it's not equal to

E {(X)}^{2}

Recall that $X^{2}$ is simply the composition of the function $X$ and the function $\cdot^{2}$ , denote $Y : = X^{2}$ so that $Y$ may only take on values $1$ and $4$ and also $P (Y = 4) = P ((X = 2) \cup (X = - 2)) = P (X = 2) + P (X = - 2) = 0.2 + 0.3 = 0.5$ and for all other values $P (Y = a) = 0$ $E (X^{2}) = E (Y) = 4 \cdot 0.5 + 1 \cdot 0.5 = 2 + .5 = 2.5$ on the other hand $E (X) = - 2 \cdot 0.2 + 1 \cdot 0.5 + 2 \cdot 0.3 = 0.7$ so that $E {(X)}^{2} = 0.49$ so $E {(X)}^{2} \neq E (X^{2})$

Expectation is Linear

Suppose that

X

is a random variable and that

a, b \in ℝ

then

a X + b

is also a random varaible and we have that

E (a X + b) = a E (X) + b

Variance

Let

X

be a random varaible and define

μ : = E (X)

then we define

Var (X) : = E ({(X - μ)}^{2})

Variance as Squares of Expection

Var (X) = E (X^{2}) - E {(X)}^{2}

Variance is Not Linear

Var (a X + b) = a^{2} X

Standard Deviation

Given a random varaible

X

we define the standard deviation of that random variable to be

σ (X) : = \sqrt{Var (X)}