Question

Suppose that Y1 and Y2 are uniformly distributed over the triangle shaded in the accompanying figure. V2 (0, 1) (-1,0) (1, 0) yi (a) Find the marginal density function for Y1 f1(91) = where کلا کا Find the marginal density function for Y2. f26/2) = where SY2 1 1 (b) (b) Find P(x2 >1/12 = }) Y1 = (Enter your probability as a fraction.) 3

Answer 1

The marginal density functions for Y1 and Y2 are both 1 since they are uniformly distributed over the shaded triangle. The probability of Y2 being greater than 1/12 is 1/4.

Step-by-step explanation:

Answer:

(a) To find the marginal density function for Y1, we need to find the probability density function (PDF) of Y1. Since Y1 is uniformly distributed over the shaded triangle, we can find the area of the triangle to obtain the PDF. The area of a triangle is given by (base x height) / 2. In this case, the base is 2 and the height is 1, so the area is 1. Therefore, the PDF for Y1 is f1(y1) = 1/1 = 1.

(b) To find the marginal density function for Y2, we can use the same approach as in part (a). The area of the triangle is again 1, so the PDF for Y2 is f2(y2) = 1/1 = 1.

(c) To find P(Y2 > 1/12), we need to calculate the probability that Y2 is greater than 1/12. Since Y2 is uniformly distributed over the shaded triangle, the probability is equal to the proportion of the triangle's area that lies above 1/12 on the y-axis. The area above 1/12 is (1/2)(1/2) = 1/4, so P(Y2 > 1/12) = 1/4.