Question #218440

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)  


1
Expert's answer
2021-07-19T05:41:23-0400

(i)


f(a,b)dbda=0302a2bxdbda\displaystyle\int_{-\infin}^{\infin}\displaystyle\int_{-\infin}^{\infin}f(a,b)dbda=\displaystyle\int_{0}^{3}\displaystyle\int_{0}^{2}a^2bxdbda

=x03a2[b22]20da=2x03a2da=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[a33]30=18x=1=>x=118=2x[\dfrac{a^3}{3}]\begin{matrix} 3 \\ 0 \end{matrix}=18x=1=>x=\dfrac{1}{18}


f(a,b)={118a2b,0a3,0b20,otherwisef(a,b) = \begin{cases} \dfrac{1}{18}a^2b, &0\leq a\leq 3, 0\leq b\leq 2 \\ 0, & otherwise \end{cases}

(ii)


fB(b)=f(a,b)da=03118a2bdaf_B(b)=\displaystyle\int_{-\infin}^{\infin}f(a,b)da=\displaystyle\int_{0}^{3} \dfrac{1}{18}a^2bda

=118b[a33]30=12b= \dfrac{1}{18}b[\dfrac{a^3}{3}]\begin{matrix} 3 \\ 0 \end{matrix}=\dfrac{1}{2}b

fB(b)={12b,0b20,otherwisef_B(b) = \begin{cases} \dfrac{1}{2}b, & 0\leq b\leq 2 \\ 0, & otherwise \end{cases}

(iii) For 0b20\leq b\leq 2 we obtain


f(ab)=f(a,b)fB(b)=19a2,0a3f(a|b)=\dfrac{f(a, b)}{f_B(b)}=\dfrac{1}{9}a^2,0\leq a\leq 3

f(a<1b=1)=0119a2da=127f(a<1|b=1)=\displaystyle\int_{0}^{1} \dfrac{1}{9}a^2da=\dfrac{1}{27}


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