Answer to Question #218440 in Statistics and Probability for Nurul Ain

Question

Let A and B are the bivariate continuous random variables. The joint probability density function is given by 𝑓(𝑎, 𝑏) = { 𝑎 2𝑏𝑥 0 < 𝑎 < 3, 0 < 𝑏 < 2; 0 otherwise. (i) Determine the value of x for which the function can serve as a joint probability density function. (ii) Find the marginal density function of B. (iii) Find 𝑓(𝑎 < 1|𝑏 = 1)  

Answer 1

(i)

$$\displaystyle\int_{-\infin}^{\infin}\displaystyle\int_{-\infin}^{\infin}f(a,b)dbda=\displaystyle\int_{0}^{3}\displaystyle\int_{0}^{2}a^2bxdbda$$

$$=x\displaystyle\int_{0}^{3}a^2[\dfrac{b^2}{2}]\begin{matrix} 2 \\ 0 \end{matrix}da=2x\displaystyle\int_{0}^{3}a^2da$$

$$=2x[\dfrac{a^3}{3}]\begin{matrix} 3 \\ 0 \end{matrix}=18x=1=>x=\dfrac{1}{18}$$

$$f(a,b) = \begin{cases} \dfrac{1}{18}a^2b, &0\leq a\leq 3, 0\leq b\leq 2 \\ 0, & otherwise \end{cases}$$

(ii)

$$f_B(b)=\displaystyle\int_{-\infin}^{\infin}f(a,b)da=\displaystyle\int_{0}^{3} \dfrac{1}{18}a^2bda$$

$$= \dfrac{1}{18}b[\dfrac{a^3}{3}]\begin{matrix} 3 \\ 0 \end{matrix}=\dfrac{1}{2}b$$

$$f_B(b) = \begin{cases} \dfrac{1}{2}b, & 0\leq b\leq 2 \\ 0, & otherwise \end{cases}$$

(iii) For $0\leq b\leq 2$ we obtain

$$f(a|b)=\dfrac{f(a, b)}{f_B(b)}=\dfrac{1}{9}a^2,0\leq a\leq 3$$

$$f(a<1|b=1)=\displaystyle\int_{0}^{1} \dfrac{1}{9}a^2da=\dfrac{1}{27}$$