Let X be a discrete random variable such that E(X)=3 and V(X)=5. Let 𝑌 = 2𝑋^2 − 3𝑋. Determine E(Y).



Answer to a math question Let X be a discrete random variable such that E(X)=3 and V(X)=5. Let 𝑌 = 2𝑋^2 − 3𝑋. Determine E(Y).

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Para encontrar E(Y), necesitamos encontrar la media de la variable aleatoria Y.

Dado que Y está definida en función de X, podemos usar las propiedades de la esperanza para simplificar el cálculo.

Primero, vamos a expandir la expresión para Y:

Y = 2X^2 - 3X

A continuación, calculamos la esperanza de Y utilizando la linealidad de la esperanza:

E(Y) = E(2X^2) - E(3X)

Sabemos que la esperanza es un operador lineal, por lo que podemos sacar las constantes fuera de la expectación:

E(Y) = 2E(X^2) - 3E(X)

Finalmente, necesitamos utilizar la información proporcionada sobre X para calcular E(X^2). La fórmula de la varianza nos dice que:

V(X) = E(X^2) - (E(X))^2

Podemos reorganizar esta fórmula para obtener E(X^2) en términos de V(X):

E(X^2) = V(X) + (E(X))^2

Sustituyendo los valores conocidos:

E(X^2) = 5 + 3^2 = 5 + 9 = 14

Ahora podemos retornar a la expresión para E(Y):

E(Y) = 2E(X^2) - 3E(X) = 2(14) - 3(3) = 28 - 9 = 19

Por lo tanto, la esperanza de la variable aleatoria Y es 19.

\textbf{Answer:} La esperanza de Y, E(Y), es 19.

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