To find E[(2X+3Y)^2], we first need to expand the square. This gives us:

E[(2X+3Y)^2] = E[4X^2 + 12XY + 9Y^2]

Since X and Y are independent, E[XY] = E[X]E[Y]. We can substitute this into the equation:

E[(2X+3Y)^2] = E[4X^2] + E[12XY] + E[9Y^2]
= 4E[X^2] + 12E[X]E[Y] + 9E[Y^2]

We know that E[X^2] = Var(X) + (E[X])^2 and E[Y^2] = Var(Y) + (E[Y])^2. Substituting these values in gives us:

E[(2X+3Y)^2] = 4(Var(X) + (E[X])^2) + 12E[X]E[Y] + 9(Var(Y) + (E[Y])^2)
= 4(2 + 1^2) + 12*1*3 + 9(4 + 3^2)
= 4*3 + 36 + 9*13
= 12 + 36 + 117
= 165

Question

To find E[(2X+3Y)^2], we first need to expand the square. This gives us:

E[(2X+3Y)^2] = E[4X^2 + 12XY + 9Y^2]

Since X and Y are independent, E[XY] = E[X]E[Y]. We can substitute this into the equation:

E[(2X+3Y)^2] = E[4X^2] + E[12XY] + E[9Y^2]
              = 4E[X^2] + 12E[X]E[Y] + 9E[Y^2]

We know that E[X^2] = Var(X) + (E[X])^2 and E[Y^2] = Var(Y) + (E[Y])^2. Substituting these values in gives us:

E[(2X+3Y)^2] = 4(Var(X) + (E[X])^2) + 12E[X]E[Y] + 9(Var(Y) + (E[Y])^2)
              = 4(2 + 1^2) + 12*1*3 + 9(4 + 3^2)
              = 4*3 + 36 + 9*13
              = 12 + 36 + 117
              = 165

Knowee AI · Accepted Answer

To find E[(2X+3Y)^2], we first need to expand the square. This gives us:

E[(2X+3Y)^2] = E[4X^2 + 12XY + 9Y^2]

Since X and Y are independent, E[XY] = E[X]E[Y]. We can substitute this into the equation:

E[(2X+3Y)^2] = E[4X^2] + E[12XY] + E[9Y^2]
              = 4E[X^2] + 12E[X]E[Y] + 9E[Y^2]

We know that E[X^2] = Var(X) + (E[X])^2 and E[Y^2] = Var(Y) + (E[Y])^2. Substituting these values in gives us:

E[(2X+3Y)^2] = 4(Var(X) + (E[X])^2) + 12E[X]E[Y] + 9(Var(Y) + (E[Y])^2)
              = 4(2 + 1^2) + 12*1*3 + 9(4 + 3^2)
              = 4*3 + 36 + 9*13
              = 12 + 36 + 117
              = 165

If 𝑋X and 𝑌Y are independent random variables with 𝐸[𝑋]=1E[X]=1, Var(𝑋)=2Var(X)=2, 𝐸[𝑌]=3E[Y]=3, and Var(𝑌)Var(Y) = 4, then find 𝐸[(2𝑋+3𝑌)2]E[(2X+3Y) 2 ].

Question

Solution

Similar Questions

Upgrade your grade with Knowee