Finding expected perimeter and area of rectangle using random distribution

Question

Hi i have a question: A random rectangle is formed in the following way- Base X is generated from Gamma(1, λ) distribution and after generating the base, the height Y is chosen uniformly on the interval [0,X]. It asks to find the expected perimeter and expected area of the random rectangle.

My approach for perimeter: E(2(X+H))= E[E(2(X+H)|X)]

                                    = E[2E(X|X) +2E(H|X)

                                    = E[2X + 2 E(H)]

                                    = 1 + 2/λ

        for area:    E(XH) = E[E(XH|X)]

                           = E[XE(H|X)]

                           = ....

Could you let me know if my approach to perimeter is right and also give me ideas as to how to approach the next steps for the area?

Thank you,

Answer 1

The base is given by $X$, the height by $\Theta X$ where $X \sim \Gamma(1,\lambda)$ and $\Theta \sim \text{unif}(0,1)$.

Presumably $X$ and $\Theta$ are independent.

Then the perimeter is $P = 2X + 2 \Theta X$, and $EP = 2 E X + 2 E \Theta E X = 2\frac{1}{\lambda}+ 2 \frac{1}{2} \frac{1}{\lambda} = \frac{3}{\lambda}$.

The area is $A = \Theta X^2$, and $E A = E \Theta E X^2 = E \Theta (\operatorname{Var}X+(E X)^2) = \frac{1}{2}(\frac{1}{\lambda^2}+\frac{1}{\lambda^2}) = \frac{1}{\lambda^2}$

Answer 2

Conditional on $X$, height is uniformly distributed between $0$ and $X$ so its expectation is $X/2$. So expected value of perimeter should be $E[2X+2E[Y|X]]=E[2X+2(X/2)]=3E[X]$. Similarly, expected value of area should be expected value of $E[X^2/2]=(Var[X]+E[X]^2)/2$.