Continuous Random Variable X, Find pdf of Y when Y = √X. Part d. [closed]











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A continuous random variable X has the following probability density function:
􏰂 k(x+x3), 0≤x≤2 f(x) = 0, otherwise
(a) Find the value of k
(b) Find the mean and variance of X,
(c) Find P(X ≤ 1).
(d) If Y =√X, determine the probability density function for Y, and hence, or otherwise, find the mean and variance for Y.



I am struggling with part d. I am not sure what to do as this is a non-linear relationship between X and Y.










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closed as off-topic by José Carlos Santos, NCh, amWhy, Shailesh, John B Dec 1 at 0:44


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, NCh, amWhy, Shailesh, John B

If this question can be reworded to fit the rules in the help center, please edit the question.













  • refer math.meta.stackexchange.com/questions/5020/… for proper mathjax formatting
    – idea
    Nov 23 at 18:16















up vote
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A continuous random variable X has the following probability density function:
􏰂 k(x+x3), 0≤x≤2 f(x) = 0, otherwise
(a) Find the value of k
(b) Find the mean and variance of X,
(c) Find P(X ≤ 1).
(d) If Y =√X, determine the probability density function for Y, and hence, or otherwise, find the mean and variance for Y.



I am struggling with part d. I am not sure what to do as this is a non-linear relationship between X and Y.










share|cite|improve this question













closed as off-topic by José Carlos Santos, NCh, amWhy, Shailesh, John B Dec 1 at 0:44


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, NCh, amWhy, Shailesh, John B

If this question can be reworded to fit the rules in the help center, please edit the question.













  • refer math.meta.stackexchange.com/questions/5020/… for proper mathjax formatting
    – idea
    Nov 23 at 18:16













up vote
-1
down vote

favorite









up vote
-1
down vote

favorite











A continuous random variable X has the following probability density function:
􏰂 k(x+x3), 0≤x≤2 f(x) = 0, otherwise
(a) Find the value of k
(b) Find the mean and variance of X,
(c) Find P(X ≤ 1).
(d) If Y =√X, determine the probability density function for Y, and hence, or otherwise, find the mean and variance for Y.



I am struggling with part d. I am not sure what to do as this is a non-linear relationship between X and Y.










share|cite|improve this question













A continuous random variable X has the following probability density function:
􏰂 k(x+x3), 0≤x≤2 f(x) = 0, otherwise
(a) Find the value of k
(b) Find the mean and variance of X,
(c) Find P(X ≤ 1).
(d) If Y =√X, determine the probability density function for Y, and hence, or otherwise, find the mean and variance for Y.



I am struggling with part d. I am not sure what to do as this is a non-linear relationship between X and Y.







probability-theory probability-distributions problem-solving






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asked Nov 23 at 17:27









J Borland

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11




closed as off-topic by José Carlos Santos, NCh, amWhy, Shailesh, John B Dec 1 at 0:44


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, NCh, amWhy, Shailesh, John B

If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by José Carlos Santos, NCh, amWhy, Shailesh, John B Dec 1 at 0:44


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, NCh, amWhy, Shailesh, John B

If this question can be reworded to fit the rules in the help center, please edit the question.












  • refer math.meta.stackexchange.com/questions/5020/… for proper mathjax formatting
    – idea
    Nov 23 at 18:16


















  • refer math.meta.stackexchange.com/questions/5020/… for proper mathjax formatting
    – idea
    Nov 23 at 18:16
















refer math.meta.stackexchange.com/questions/5020/… for proper mathjax formatting
– idea
Nov 23 at 18:16




refer math.meta.stackexchange.com/questions/5020/… for proper mathjax formatting
– idea
Nov 23 at 18:16










2 Answers
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For (d) $F_Y(y)=P(Yle y)=P(sqrt{X}le y)=P(Xle y^2)={kint_0^{y^2}(x+x^3)dx=k(frac{y^4}{2}+frac{y^8}{4})}$



The density function $f_Y(y)=k(2y^3+2y^7)$, for $0le ylesqrt{2}$.



I'll leave the mean and variance calculation to you.






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    We have $$Pr(Y<y)=Pr{(sqrt X<y)}=Pr(X<y^2)=F_X(y^2)$$therefore$$f_Y(y)=2yf_X(y^2)$$






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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

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      active

      oldest

      votes








      up vote
      0
      down vote













      For (d) $F_Y(y)=P(Yle y)=P(sqrt{X}le y)=P(Xle y^2)={kint_0^{y^2}(x+x^3)dx=k(frac{y^4}{2}+frac{y^8}{4})}$



      The density function $f_Y(y)=k(2y^3+2y^7)$, for $0le ylesqrt{2}$.



      I'll leave the mean and variance calculation to you.






      share|cite|improve this answer

























        up vote
        0
        down vote













        For (d) $F_Y(y)=P(Yle y)=P(sqrt{X}le y)=P(Xle y^2)={kint_0^{y^2}(x+x^3)dx=k(frac{y^4}{2}+frac{y^8}{4})}$



        The density function $f_Y(y)=k(2y^3+2y^7)$, for $0le ylesqrt{2}$.



        I'll leave the mean and variance calculation to you.






        share|cite|improve this answer























          up vote
          0
          down vote










          up vote
          0
          down vote









          For (d) $F_Y(y)=P(Yle y)=P(sqrt{X}le y)=P(Xle y^2)={kint_0^{y^2}(x+x^3)dx=k(frac{y^4}{2}+frac{y^8}{4})}$



          The density function $f_Y(y)=k(2y^3+2y^7)$, for $0le ylesqrt{2}$.



          I'll leave the mean and variance calculation to you.






          share|cite|improve this answer












          For (d) $F_Y(y)=P(Yle y)=P(sqrt{X}le y)=P(Xle y^2)={kint_0^{y^2}(x+x^3)dx=k(frac{y^4}{2}+frac{y^8}{4})}$



          The density function $f_Y(y)=k(2y^3+2y^7)$, for $0le ylesqrt{2}$.



          I'll leave the mean and variance calculation to you.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 23 at 18:37









          herb steinberg

          2,2932310




          2,2932310






















              up vote
              0
              down vote













              We have $$Pr(Y<y)=Pr{(sqrt X<y)}=Pr(X<y^2)=F_X(y^2)$$therefore$$f_Y(y)=2yf_X(y^2)$$






              share|cite|improve this answer

























                up vote
                0
                down vote













                We have $$Pr(Y<y)=Pr{(sqrt X<y)}=Pr(X<y^2)=F_X(y^2)$$therefore$$f_Y(y)=2yf_X(y^2)$$






                share|cite|improve this answer























                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  We have $$Pr(Y<y)=Pr{(sqrt X<y)}=Pr(X<y^2)=F_X(y^2)$$therefore$$f_Y(y)=2yf_X(y^2)$$






                  share|cite|improve this answer












                  We have $$Pr(Y<y)=Pr{(sqrt X<y)}=Pr(X<y^2)=F_X(y^2)$$therefore$$f_Y(y)=2yf_X(y^2)$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 23 at 18:56









                  Mostafa Ayaz

                  13.3k3836




                  13.3k3836















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