2022-04-26

star233

To find the marginal probability density functions of X and Y, we need to integrate the joint probability density function f(x,y) over the other variable.
First, let's find the marginal probability density function of X:
${f}_{X}\left(x\right)={\int }_{-\infty }^{\infty }f\left(x,y\right)dy$
Since f(x,y) is zero for y outside the range [0,1], we can rewrite the above integral as:
${f}_{X}\left(x\right)={\int }_{0}^{1}f\left(x,y\right)dy$
Using the definition of f(x,y), we have:
${f}_{X}\left(x\right)={\int }_{0}^{1}\frac{2}{3}\left(x+2y\right)dy$
Evaluating the integral, we get:
${f}_{X}\left(x\right)=\frac{2}{3}x+\frac{1}{3}$
for $0\le x\le 1$, and ${f}_{X}\left(x\right)=0$ elsewhere.
Next, let's find the marginal probability density function of Y:
${f}_{Y}\left(y\right)={\int }_{-\infty }^{\infty }f\left(x,y\right)dx$
Again, since f(x,y) is zero for x outside the range [0,1], we can rewrite the above integral as:
${f}_{Y}\left(y\right)={\int }_{0}^{1}f\left(x,y\right)dx$
Using the definition of f(x,y), we have:
${f}_{Y}\left(y\right)={\int }_{0}^{1}\frac{2}{3}\left(x+2y\right)dx$
Evaluating the integral, we get:
${f}_{Y}\left(y\right)=\frac{4}{3}y+\frac{1}{3}$
for $0\le y\le 1$, and ${f}_{Y}\left(y\right)=0$ elsewhere.
Therefore, the marginal probability density function of X is:
${f}_{X}\left(x\right)=\left\{\begin{array}{cc}\frac{2}{3}x+\frac{1}{3}\hfill & 0\le x\le 1\hfill \\ 0\hfill & \text{elsewhere}\hfill \end{array}$
and the marginal probability density function of Y is:
${f}_{Y}\left(y\right)=\left\{\begin{array}{cc}\frac{4}{3}y+\frac{1}{3}\hfill & 0\le y\le 1\hfill \\ 0\hfill & \text{elsewhere}\hfill \end{array}$

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