Random Variable X and Y has a joint probability density function. Find $f_ Y(x | y)$

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Random Variable X and Y has a joint probability density function.



$$f_X, Y (x, y) =begincases
c(x + 3y)& 5 leq x leq y, 6 leq y leq10\
0 & textotherwise \
endcases
$$



(a) Find $f_X (x | y)$



(b) $P(x leq 5 | Y = 9)$



My attempt:



$f_X (x | y) = fracf_X, Y(x, y)f_Y(y)$



$$f_Y(y) = cint_5^y(x+3y)dx = c/2 (-25 - 30 y + 7 y^2)$$



for $f_Y(y)$ has support $6 leq y leq 10$, 0 otherwise



$$f_X (x | y) = frac0.5c(x+3y)(-25 - 30 y + 7 y^2)$$



$f_X (x | y)$ has support the same as the joint probability function



(b)



$f_X (x | y = 9) = fracx+27272$



$$P(x leq 5 | Y = 9) = int_5^? f_X (x | y = 9)dx = $$



Not sure




probability




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  • 1




    Replace $?$ by $5$ and the answer is $0$.
    – Piyush Divyanakar
    Jul 19 at 10:15














up vote
1
down vote

favorite












Random Variable X and Y has a joint probability density function.



$$f_X, Y (x, y) =begincases
c(x + 3y)& 5 leq x leq y, 6 leq y leq10\
0 & textotherwise \
endcases
$$



(a) Find $f_X (x | y)$



(b) $P(x leq 5 | Y = 9)$



My attempt:



$f_X (x | y) = fracf_X, Y(x, y)f_Y(y)$



$$f_Y(y) = cint_5^y(x+3y)dx = c/2 (-25 - 30 y + 7 y^2)$$



for $f_Y(y)$ has support $6 leq y leq 10$, 0 otherwise



$$f_X (x | y) = frac0.5c(x+3y)(-25 - 30 y + 7 y^2)$$



$f_X (x | y)$ has support the same as the joint probability function



(b)



$f_X (x | y = 9) = fracx+27272$



$$P(x leq 5 | Y = 9) = int_5^? f_X (x | y = 9)dx = $$



Not sure




probability




share|cite|improve this question















  • 1




    Replace $?$ by $5$ and the answer is $0$.
    – Piyush Divyanakar
    Jul 19 at 10:15












up vote
1
down vote

favorite









up vote
1
down vote

favorite











Random Variable X and Y has a joint probability density function.



$$f_X, Y (x, y) =begincases
c(x + 3y)& 5 leq x leq y, 6 leq y leq10\
0 & textotherwise \
endcases
$$



(a) Find $f_X (x | y)$



(b) $P(x leq 5 | Y = 9)$



My attempt:



$f_X (x | y) = fracf_X, Y(x, y)f_Y(y)$



$$f_Y(y) = cint_5^y(x+3y)dx = c/2 (-25 - 30 y + 7 y^2)$$



for $f_Y(y)$ has support $6 leq y leq 10$, 0 otherwise



$$f_X (x | y) = frac0.5c(x+3y)(-25 - 30 y + 7 y^2)$$



$f_X (x | y)$ has support the same as the joint probability function



(b)



$f_X (x | y = 9) = fracx+27272$



$$P(x leq 5 | Y = 9) = int_5^? f_X (x | y = 9)dx = $$



Not sure




probability




share|cite|improve this question











Random Variable X and Y has a joint probability density function.



$$f_X, Y (x, y) =begincases
c(x + 3y)& 5 leq x leq y, 6 leq y leq10\
0 & textotherwise \
endcases
$$



(a) Find $f_X (x | y)$



(b) $P(x leq 5 | Y = 9)$



My attempt:



$f_X (x | y) = fracf_X, Y(x, y)f_Y(y)$



$$f_Y(y) = cint_5^y(x+3y)dx = c/2 (-25 - 30 y + 7 y^2)$$



for $f_Y(y)$ has support $6 leq y leq 10$, 0 otherwise



$$f_X (x | y) = frac0.5c(x+3y)(-25 - 30 y + 7 y^2)$$



$f_X (x | y)$ has support the same as the joint probability function



(b)



$f_X (x | y = 9) = fracx+27272$



$$P(x leq 5 | Y = 9) = int_5^? f_X (x | y = 9)dx = $$



Not sure





probability





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asked Jul 19 at 7:39









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  • 1




    Replace $?$ by $5$ and the answer is $0$.
    – Piyush Divyanakar
    Jul 19 at 10:15












  • 1




    Replace $?$ by $5$ and the answer is $0$.
    – Piyush Divyanakar
    Jul 19 at 10:15







1




1




Replace $?$ by $5$ and the answer is $0$.
– Piyush Divyanakar
Jul 19 at 10:15




Replace $?$ by $5$ and the answer is $0$.
– Piyush Divyanakar
Jul 19 at 10:15










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(a) Everything is done okay, save that the final answer should be $$f_Xmid Y(xmid y)=dfrac2(x+3y)7y^2-30y-25mathbf 1_5leqslant xleqslant y, 6leqslant yleqslant 10$$



(b) Whatever value of $Y$, the support for $X$ indicates that it is impossible to have $Xleqslant 5$. $$mathsf P(Xleqslant5mid Y=9)=0$$






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    up vote
    0
    down vote



    accepted










    (a) Everything is done okay, save that the final answer should be $$f_Xmid Y(xmid y)=dfrac2(x+3y)7y^2-30y-25mathbf 1_5leqslant xleqslant y, 6leqslant yleqslant 10$$



    (b) Whatever value of $Y$, the support for $X$ indicates that it is impossible to have $Xleqslant 5$. $$mathsf P(Xleqslant5mid Y=9)=0$$






    share|cite|improve this answer



























      up vote
      0
      down vote



      accepted










      (a) Everything is done okay, save that the final answer should be $$f_Xmid Y(xmid y)=dfrac2(x+3y)7y^2-30y-25mathbf 1_5leqslant xleqslant y, 6leqslant yleqslant 10$$



      (b) Whatever value of $Y$, the support for $X$ indicates that it is impossible to have $Xleqslant 5$. $$mathsf P(Xleqslant5mid Y=9)=0$$






      share|cite|improve this answer

























        up vote
        0
        down vote



        accepted







        up vote
        0
        down vote



        accepted






        (a) Everything is done okay, save that the final answer should be $$f_Xmid Y(xmid y)=dfrac2(x+3y)7y^2-30y-25mathbf 1_5leqslant xleqslant y, 6leqslant yleqslant 10$$



        (b) Whatever value of $Y$, the support for $X$ indicates that it is impossible to have $Xleqslant 5$. $$mathsf P(Xleqslant5mid Y=9)=0$$






        share|cite|improve this answer















        (a) Everything is done okay, save that the final answer should be $$f_Xmid Y(xmid y)=dfrac2(x+3y)7y^2-30y-25mathbf 1_5leqslant xleqslant y, 6leqslant yleqslant 10$$



        (b) Whatever value of $Y$, the support for $X$ indicates that it is impossible to have $Xleqslant 5$. $$mathsf P(Xleqslant5mid Y=9)=0$$







        share|cite|improve this answer















        share|cite|improve this answer



        share|cite|improve this answer








        answered Jul 19 at 10:46



























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