Probability - point inside disk












0












$begingroup$



$X, Y sim mathcal{N}(0,1)$ are independent random variables. Let $D_r$ be the disk centered at the origin with radius $r$. Find $r$ such that $mathbb{P}[(X,Y) in D_r ] = 0.3$.




My attempt: $(X, Y) in D_r$ means $X^2 + Y^2 le r^2$. What should I do next?






probability normal-distribution







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  • 4




    $begingroup$
    Yours is not really an attempt at all. Maybe do some integration? What is the law of the couple $(X,Y)$?
    $endgroup$
    – Federico
    Dec 10 '18 at 16:07


















0












$begingroup$



$X, Y sim mathcal{N}(0,1)$ are independent random variables. Let $D_r$ be the disk centered at the origin with radius $r$. Find $r$ such that $mathbb{P}[(X,Y) in D_r ] = 0.3$.




My attempt: $(X, Y) in D_r$ means $X^2 + Y^2 le r^2$. What should I do next?






probability normal-distribution







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    Yours is not really an attempt at all. Maybe do some integration? What is the law of the couple $(X,Y)$?
    $endgroup$
    – Federico
    Dec 10 '18 at 16:07
















0












0








0





$begingroup$



$X, Y sim mathcal{N}(0,1)$ are independent random variables. Let $D_r$ be the disk centered at the origin with radius $r$. Find $r$ such that $mathbb{P}[(X,Y) in D_r ] = 0.3$.




My attempt: $(X, Y) in D_r$ means $X^2 + Y^2 le r^2$. What should I do next?






probability normal-distribution







share|cite|improve this question











$endgroup$





$X, Y sim mathcal{N}(0,1)$ are independent random variables. Let $D_r$ be the disk centered at the origin with radius $r$. Find $r$ such that $mathbb{P}[(X,Y) in D_r ] = 0.3$.




My attempt: $(X, Y) in D_r$ means $X^2 + Y^2 le r^2$. What should I do next?






probability normal-distribution




probability normal-distribution






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 10 '18 at 16:12









gt6989b

35k22557




35k22557










asked Dec 10 '18 at 16:05









AlphaDelphiAlphaDelphi

11




11








  • 4




    $begingroup$
    Yours is not really an attempt at all. Maybe do some integration? What is the law of the couple $(X,Y)$?
    $endgroup$
    – Federico
    Dec 10 '18 at 16:07
















  • 4




    $begingroup$
    Yours is not really an attempt at all. Maybe do some integration? What is the law of the couple $(X,Y)$?
    $endgroup$
    – Federico
    Dec 10 '18 at 16:07










4




4




$begingroup$
Yours is not really an attempt at all. Maybe do some integration? What is the law of the couple $(X,Y)$?
$endgroup$
– Federico
Dec 10 '18 at 16:07






$begingroup$
Yours is not really an attempt at all. Maybe do some integration? What is the law of the couple $(X,Y)$?
$endgroup$
– Federico
Dec 10 '18 at 16:07












2 Answers
2






active

oldest

votes


















1












$begingroup$

HINT



For any event $A$, you have
$$
mathbb{P}[A] = iint_{mathbb{R}^2} mathbb{I}_{A(x,y)}f(x,y)dxdy,
$$
and in your case, since $X$ and $Y$ are independent, $f(x,y) = phi(x) phi(y)$.



Convert the integration over the plane with an indicator function to integrating over a region, where the indicator is always true, and compute the probability...






share|cite|improve this answer









$endgroup$





















    -1












    $begingroup$

    The pdf of $X$ is $f_X(x)=frac{e^{-x^2/2}}{sqrt{2pi}}$. Same for $Y$. The pdf for $(X,Y)$ is $f_{X,Y}(x,y)=f_X(x)f_Y(y) = frac{e^{-(x^2+y^2)/2}}{2pi}$. In polar coordinates,
    $$
    begin{split}
    mathbb P(B_R) &=
    int_{-infty}^inftyint_{-infty}^infty mathbf 1_{x^2+y^2<R}(x,y)f_{X,Y}(x,y),dx,dy \
    &= int_{-infty}^inftyint_{-infty}^infty mathbf 1_{x^2+y^2<R}(x,y)frac{e^{-(x^2+y^2)/2}}{2pi}(x,y),dx,dy \
    &= int_0^R 2pi r frac{e^{-r^2/2}}{2pi},dr \
    &= 1-e^{R^2/2}.
    end{split}
    $$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      -1. I would strongly prefer for the OP to do some work himself without us spoonfeeding the answers to him without any work on his part
      $endgroup$
      – gt6989b
      Dec 10 '18 at 16:18










    • $begingroup$
      @gt6989b Well, I don't see that as a good reason to -1...
      $endgroup$
      – Federico
      Dec 10 '18 at 16:18











    Your Answer





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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    HINT



    For any event $A$, you have
    $$
    mathbb{P}[A] = iint_{mathbb{R}^2} mathbb{I}_{A(x,y)}f(x,y)dxdy,
    $$
    and in your case, since $X$ and $Y$ are independent, $f(x,y) = phi(x) phi(y)$.



    Convert the integration over the plane with an indicator function to integrating over a region, where the indicator is always true, and compute the probability...






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      HINT



      For any event $A$, you have
      $$
      mathbb{P}[A] = iint_{mathbb{R}^2} mathbb{I}_{A(x,y)}f(x,y)dxdy,
      $$
      and in your case, since $X$ and $Y$ are independent, $f(x,y) = phi(x) phi(y)$.



      Convert the integration over the plane with an indicator function to integrating over a region, where the indicator is always true, and compute the probability...






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        HINT



        For any event $A$, you have
        $$
        mathbb{P}[A] = iint_{mathbb{R}^2} mathbb{I}_{A(x,y)}f(x,y)dxdy,
        $$
        and in your case, since $X$ and $Y$ are independent, $f(x,y) = phi(x) phi(y)$.



        Convert the integration over the plane with an indicator function to integrating over a region, where the indicator is always true, and compute the probability...






        share|cite|improve this answer









        $endgroup$



        HINT



        For any event $A$, you have
        $$
        mathbb{P}[A] = iint_{mathbb{R}^2} mathbb{I}_{A(x,y)}f(x,y)dxdy,
        $$
        and in your case, since $X$ and $Y$ are independent, $f(x,y) = phi(x) phi(y)$.



        Convert the integration over the plane with an indicator function to integrating over a region, where the indicator is always true, and compute the probability...







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 10 '18 at 16:11









        gt6989bgt6989b

        35k22557




        35k22557























            -1












            $begingroup$

            The pdf of $X$ is $f_X(x)=frac{e^{-x^2/2}}{sqrt{2pi}}$. Same for $Y$. The pdf for $(X,Y)$ is $f_{X,Y}(x,y)=f_X(x)f_Y(y) = frac{e^{-(x^2+y^2)/2}}{2pi}$. In polar coordinates,
            $$
            begin{split}
            mathbb P(B_R) &=
            int_{-infty}^inftyint_{-infty}^infty mathbf 1_{x^2+y^2<R}(x,y)f_{X,Y}(x,y),dx,dy \
            &= int_{-infty}^inftyint_{-infty}^infty mathbf 1_{x^2+y^2<R}(x,y)frac{e^{-(x^2+y^2)/2}}{2pi}(x,y),dx,dy \
            &= int_0^R 2pi r frac{e^{-r^2/2}}{2pi},dr \
            &= 1-e^{R^2/2}.
            end{split}
            $$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              -1. I would strongly prefer for the OP to do some work himself without us spoonfeeding the answers to him without any work on his part
              $endgroup$
              – gt6989b
              Dec 10 '18 at 16:18










            • $begingroup$
              @gt6989b Well, I don't see that as a good reason to -1...
              $endgroup$
              – Federico
              Dec 10 '18 at 16:18
















            -1












            $begingroup$

            The pdf of $X$ is $f_X(x)=frac{e^{-x^2/2}}{sqrt{2pi}}$. Same for $Y$. The pdf for $(X,Y)$ is $f_{X,Y}(x,y)=f_X(x)f_Y(y) = frac{e^{-(x^2+y^2)/2}}{2pi}$. In polar coordinates,
            $$
            begin{split}
            mathbb P(B_R) &=
            int_{-infty}^inftyint_{-infty}^infty mathbf 1_{x^2+y^2<R}(x,y)f_{X,Y}(x,y),dx,dy \
            &= int_{-infty}^inftyint_{-infty}^infty mathbf 1_{x^2+y^2<R}(x,y)frac{e^{-(x^2+y^2)/2}}{2pi}(x,y),dx,dy \
            &= int_0^R 2pi r frac{e^{-r^2/2}}{2pi},dr \
            &= 1-e^{R^2/2}.
            end{split}
            $$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              -1. I would strongly prefer for the OP to do some work himself without us spoonfeeding the answers to him without any work on his part
              $endgroup$
              – gt6989b
              Dec 10 '18 at 16:18










            • $begingroup$
              @gt6989b Well, I don't see that as a good reason to -1...
              $endgroup$
              – Federico
              Dec 10 '18 at 16:18














            -1












            -1








            -1





            $begingroup$

            The pdf of $X$ is $f_X(x)=frac{e^{-x^2/2}}{sqrt{2pi}}$. Same for $Y$. The pdf for $(X,Y)$ is $f_{X,Y}(x,y)=f_X(x)f_Y(y) = frac{e^{-(x^2+y^2)/2}}{2pi}$. In polar coordinates,
            $$
            begin{split}
            mathbb P(B_R) &=
            int_{-infty}^inftyint_{-infty}^infty mathbf 1_{x^2+y^2<R}(x,y)f_{X,Y}(x,y),dx,dy \
            &= int_{-infty}^inftyint_{-infty}^infty mathbf 1_{x^2+y^2<R}(x,y)frac{e^{-(x^2+y^2)/2}}{2pi}(x,y),dx,dy \
            &= int_0^R 2pi r frac{e^{-r^2/2}}{2pi},dr \
            &= 1-e^{R^2/2}.
            end{split}
            $$






            share|cite|improve this answer









            $endgroup$



            The pdf of $X$ is $f_X(x)=frac{e^{-x^2/2}}{sqrt{2pi}}$. Same for $Y$. The pdf for $(X,Y)$ is $f_{X,Y}(x,y)=f_X(x)f_Y(y) = frac{e^{-(x^2+y^2)/2}}{2pi}$. In polar coordinates,
            $$
            begin{split}
            mathbb P(B_R) &=
            int_{-infty}^inftyint_{-infty}^infty mathbf 1_{x^2+y^2<R}(x,y)f_{X,Y}(x,y),dx,dy \
            &= int_{-infty}^inftyint_{-infty}^infty mathbf 1_{x^2+y^2<R}(x,y)frac{e^{-(x^2+y^2)/2}}{2pi}(x,y),dx,dy \
            &= int_0^R 2pi r frac{e^{-r^2/2}}{2pi},dr \
            &= 1-e^{R^2/2}.
            end{split}
            $$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 10 '18 at 16:17









            FedericoFederico

            5,144514




            5,144514












            • $begingroup$
              -1. I would strongly prefer for the OP to do some work himself without us spoonfeeding the answers to him without any work on his part
              $endgroup$
              – gt6989b
              Dec 10 '18 at 16:18










            • $begingroup$
              @gt6989b Well, I don't see that as a good reason to -1...
              $endgroup$
              – Federico
              Dec 10 '18 at 16:18


















            • $begingroup$
              -1. I would strongly prefer for the OP to do some work himself without us spoonfeeding the answers to him without any work on his part
              $endgroup$
              – gt6989b
              Dec 10 '18 at 16:18










            • $begingroup$
              @gt6989b Well, I don't see that as a good reason to -1...
              $endgroup$
              – Federico
              Dec 10 '18 at 16:18
















            $begingroup$
            -1. I would strongly prefer for the OP to do some work himself without us spoonfeeding the answers to him without any work on his part
            $endgroup$
            – gt6989b
            Dec 10 '18 at 16:18




            $begingroup$
            -1. I would strongly prefer for the OP to do some work himself without us spoonfeeding the answers to him without any work on his part
            $endgroup$
            – gt6989b
            Dec 10 '18 at 16:18












            $begingroup$
            @gt6989b Well, I don't see that as a good reason to -1...
            $endgroup$
            – Federico
            Dec 10 '18 at 16:18




            $begingroup$
            @gt6989b Well, I don't see that as a good reason to -1...
            $endgroup$
            – Federico
            Dec 10 '18 at 16:18


















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