Find $P(Y_1≤3/4,Y_2≥1/2)$ of a joint probability density function.

Question

Let Y1 and Y2 have the joint probability density function given by:

$ f (y_1, y_2) = 6(1−y_2), \text{for } 0≤y_1 ≤y_2 ≤1$

Find $P(Y_1≤3/4,Y_2≥1/2).$

Answer:

$$\int_{1/2}^{3/4}\int_{y_1}^{1}6(1− y_2 )dy_2dy_1 + \int_{1/2}^{1}\int_{1/2}^{1}6(1− y_2 )dy_1dy_2 = 7/64 + 24/64 = 31/64 $$

My problem is I don't understand the logic for the second part. We know for a fact that:

$$P(a_1 ≤ Y_1 ≤ a_2, b_1 ≤ Y_2 ≤ b_2) = \int^{b_2}_{b_1}\int^{a_2}_{a_1}6(1− y_2 )dy_1dy_2$$

Thus, I can make the following equations

$$ \int_{1/2}^{3/4}\int_{y_1}^{1}6(1− y_2 )dy_2dy_1 = P(1/2 \leq Y_2 \leq 1, 1/2 \leq Y_1 \leq 3/4 | Y_1 \leq Y_2) $$

This part I understand the logic of why we would add this integral to the expression

$$\int_{1/2}^{1}\int_{1/2}^{1}6(1− y_2 )dy_1dy_2 = P(1/2 \leq Y_2 \leq 1, 1/2 \leq Y_1 \leq 1)$$.

I don't understand the logic of how adding both expressions leads to $P(Y_1≤3/4,Y_2≥1/2).$. In fact, I don't see the logic of choosing $P(1/2 \leq Y_2 \leq 1, 1/2 \leq Y_1 \leq 1)$.

Answer 1

Hint: consider the figure below

By the specification of your joint PDF, you only want to integrate within triangle ABC. Now, within the ABC triangle, the region described by $\{(y_1,y_2):y_1\leq\frac{3}{4},y_2\geq\frac{1}{2}\}$ is the polygon DEGHB which can be decomposed into 2 bits: a rectangle DEIB and a quadrilateral EGHI.

Answer 2

No you are right! You should instead choose $$ P(1/2 \leq Y_2 \leq 1 \, , \, {\bf0 \leq Y_1 \leq 1/2)}. $$ This doesn't affect the answer since $$ \int_{1/2}^1 \int^{1/2}_{y_1=0} 6(1-y_2) \, dy_1 dy_2 = \int_{1/2}^1 \int^{1}_{y_1=1/2} 6(1-y_2) \, dy_1 dy_2. $$

Do you see why this makes sense? Hint: Draw the area you want to integrate over.