# Thread: Expected Value of a Function of a Random Variable Questions

1. ## Expected Value of a Function of a Random Variable Questions

1) A tool and die company makes castings for steel stress-monitoring gauges. Their annual profit, Q, in hundreds of thousands of dollars, can be expressed as a function of product demand, y:
Q(y) = 2(1-e^(-2y))
Suppose that the demand (in thousands) for their castings follows an exponential pdf, fy(y) = 6e^(-6y), y > 0. Find the company's expected profit.

2) The hypotenuse, Y, of an isosceles right triangle is a random variable having a uniform pdf over the interval [6,10]. Calculate the expected value of the triangle's area. Do not leave the answer as a function of a. (It shows the triangle with both smaller sides as a and the hypotenuse as Y.)

2. Originally Posted by Janu42
1) A tool and die company makes castings for steel stress-monitoring gauges. Their annual profit, Q, in hundreds of thousands of dollars, can be expressed as a function of product demand, y:
Q(y) = 2(1-e^(-2y))
Suppose that the demand (in thousands) for their castings follows an exponential pdf, fy(y) = 6e^(-6y), y > 0. Find the company's expected profit.

[snip]
$E(Q) = 2 E\left(1 - e^{-2y} \right) = 2 \int_0^{+\infty} (1 - e^{-2y}) 6 e^{-6y} \, dy = ....$

where your job is to simplify the integrand and then integrate.

3. Originally Posted by Janu42
[snip]
2) The hypotenuse, Y, of an isosceles right triangle is a random variable having a uniform pdf over the interval [6,10]. Calculate the expected value of the triangle's area. Do not leave the answer as a function of a. (It shows the triangle with both smaller sides as a and the hypotenuse as Y.)
I assume it's a right isosceles triangle since you talk about a hypotenuse.

$A = \frac{a^2}{2} = \frac{Y^2}{4}$ using Pythagoras' Theorem.

$E(A) = \frac{1}{4} E(Y^2) = \frac{1}{4} \int_{6}^{10} y^2 \cdot \frac{1}{4} \, dy = ....$

where your job is to simplify and then integrate.