Question

Give a joint density function of random variables X and Y as follows$\tiny \bold{f(x,y) = \begin{cases}x + y, \: \: \: \: \: 0 \leq x \leq1 , \: \: \: \: \: 0 \leq y \leq2 \\ \: \: \: \: \: \: \: \: 0, \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bold{otherwise}\end{cases}}$Find the marginal density of X​​

Answer 1

Answer:

Figure 5.8(a) shows RXY in the x−y plane.

The figure shows (a) RXY as well as (b) the integration region for finding P(Y<2X2) for Solved Problem 1.

To find the constant c, we write

1=∫∞−∞∫∞−∞fXY(x,y)dxdy=∫10∫1−x0cx+1dydx=∫10(cx+1)(1−x)dx=12+16c.

Thus, we conclude c=3.

We first note that RX=RY=[0,1].

fX(x)=∫∞−∞fXY(x,y)dy=∫1−x03x+1dy=(3x+1)(1−x), for x∈[0,1].

Thus, we have

fX(x)=⎧⎩⎨⎪⎪(3x+1)(1−x)00≤x≤1otherwise

Similarly, we obtain

fY(y)=∫∞−∞fXY(x,y)dx=∫1−y03x+1dx=12(1−y)(5−3y), for y∈[0,1].

Thus, we have

fY(y)=⎧⎩⎨⎪⎪12(1−y)(5−3y)00≤y≤1otherwise

To find P(Y<2X2), we need to integrate fXY(x,y) over the region shown in Figure 5.8(b). We have

P(Y<2X2)=∫∞−∞∫2x2−∞fXY(x,y)dydx=∫10∫min(2x2,1−x)03x+1dydx=∫10(3x+1)min(2x2,1−x)dx=∫1202x2(3x+1)dx+∫112(3x+1)(1−x)dx=5396.

Answer 2

To find the marginal density of X, we need to integrate the joint density function over all possible values of y:

$\rm f_X(x) = \int_{-\infty}^{\infty}f(x,y)\ dy$

However, since the joint density function is zero outside the region 0 ≤ x ≤ 1, 0 ≤ y ≤ 2, we only need to integrate over this region:

$\rm f_X(x) = \int_{0}^{2}(x + y)\ dy$

$\rm =\left[xy + \frac{1}{2}y^2\right]_{0}^{2} = 2x+2$

Therefore, the marginal density of X is:

$\rm f_X(x) = \begin{cases}2x+2, \: \: \: \: \: 0 \leq x \leq1 \\ \: \: \: \: \: \: \: \: 0, \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bold{otherwise}\end{cases}$