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Question

The hypothenuse, X, of an isosceles right triangle is a uniform random variable on 0, 1]. Calculate the expected value and the variance of the area of the triangle.

Answer 1

 

  Let t be the leg of the isosceles right triangle. Then t^2 + t^2 = x ^2 by pythagorean theorem.

   So 2*t^2 = x^2

   t^2 = x^2/2

 

   so t = square root (x^2/2) = x/square root(2)

   rationalizing the denominator results is (rad2/2)X where rad2 = square root(2)

 

  The area of the right triangle = 1/2 * Base * Height =   1/2 * rad2/2X * rad2/2X =     x/4

 

 So the probability function of the area in terms of the continuous random variable x is f(x) = x/4

 where x is UNIFORM(0,1)

 

The expected value E(x) = integral( xf(x) ) on the probability range. 

 

So the expected value of the area is integral x^2/4 on [0,1].

The indefinite integral is x^3/12 which integrates to 1/12

 

The variance is integral of (X - mean)^2 which in this case is

  (X- 1/2)^2.

The indefinite integral is 1/3*(x-1/2)^3 on [0,1]

 

which integrates to 1/3* (1/2)^3 - 1/3*(-1/2)^3 = 1/24 - -1/24 = 2/24 = 1/12