Econ7630_Exam1a

P[X = x] >= 0

Sum[f(x)] = 1

A function that assigns a unique numerical value to each sample space outcome.



0.1

P[X = x, Y = y]

f(x) = sum_y[f(x,y)]

f(y) = sum_x[f(x,y)]

= P[X = x, Y = y]/P[Y = y] = f(x, y)/f(y)

The conditional probabilty that X equals x given that Y equals y is equal to the joint PDF of X and Y divided by the marginal PDF of y.

(1) f(x, y) = f(x)f(y) for all x, y.

• Or (derived from (1))
• (2) f(x|y) = f(x),
• (3) f(y|x) = f(y) for all x, y.

• We know that
• (i) X,Y are independent if f(x,y) = f(x)f(y) for all x,y.
• (ii) f(x|y) = f(x,y)/f(y)

So by (i) f(x|y) = f(x)f(y)/f(y) -> f(x|y) = f(x).

The average value of X on an infinite number of experimental trials.

For example, E[X] for a six sided fair die is 3.5. Clearly saying something like the likely value of a die role is 3.5 does not make sense. Average makes sense.

• 
• Remember: even though X is being transformed by g(X), you still use the original PDF f(x).

E[c] = c

Remember: The Expected Value of a constant is that constant. This allows you to pull constants out of the E[ ] operator.

Let g(X) = c -> E[g(X)] = integral[g(x)f(x)dx] = integral[cf(x)dx] = c*integral[f(x)dx] = c(1) = c

E[aX + b] = aE[X] + b

Let g(X) = aX + b -> E[g(X)] = E[aX + b] = E[aX] + E[b] -> aE[X] + b

• 
• Remember: the expected value of the sum is the sum of the expected values.

= E[(X - mu)²] = E[X²] - mu², where mu = E[X].

• Proof:
• E[(X - mu)²] = E[(X - mu)(X - mu)] = E[X² - 2muX + mu²] = E[X²] - 2muE[X] + mu² = E[X²] - 2mu*mu + mu² = E[X²] - 2mu² + mu² = E[X²] - mu²

= b²Var[X]

• Proof:
• Let Y = a + bX -> E[Y] = a + bE[Y]. Then Var(Y) =
• 

• (x - mu)/sig
• where sig = sqrt[Var(X)] (i.e. the standard deviation)

• E[z] = E[(1/sig)X - mu/sig]
• = (1/sig)E[X] - E[mu/sig]
• = mu/sig - mu/sig = 0

• Var(z) = Var[(1/sig)X - mu/sig]
• = (1/sig)²Var(X) = sig²/sig² = 1

Remember: to get to line two, Var(aX +- b) = a²Var(X).



• = E[X₁, X₂] - mu₁mu₂
• Derivation:
• 

• 1. (X, Y) pairs tend to be both greater than their means or both less than their means.
• 2. (X, Y) pairs tend to be mixed about their means (one greater and one less)
• 3. (X, Y) pairs "evenly" spread about their means.



• 1. perfect positive relation
• 2. perfect negative relation
• 3. no linear relationship

Remember: absolute_value(p) measure the strength of the linear relationship.

= mu₁mu₂

• Proof:
• 

= 0

Remember: the converse is not true.

• Proof:
• We know E[X, Y] = mu_Xmu_Y, when X, Y ind.

cov[X, Y] = E[X, Y]-mu_Xmu_Y = mu_Xmu_Y-mu_Xmu_Y = 0

= cE[X] + dE[Y] = c*mu_X + d*mu_Y

• c₁²Var(X₁) + c₂²Var(X₂) + 2c₁c₂Cov(X₁, X₂)
• Proof:
• Var(c₁X₁ + C₂X₂)
• 

• = integral(x*f(x|y)dx)
• Note: f(x|y) = f(x,y)/f(y)

= a + b*mu

= E[g(X)]

• = E_x[E_y(Y|X)]
• Proof:
• 

• = 0
• Proof:
• 

= E[X²|Y] - E[X|Y]²



Y~N((a*mu +b), a²sig²)

cov(X,Y) = 0 <=> X,Y are independent.