Probability X1 ≥ X2












3












$begingroup$


Suppose X1 and X2 are independent Geometric p random variables. What is the
probability that X1 ≥ X2?



I am confused about this question because we aren't told anything about X1 and X2 other than they are Geometric. Wouldn't this be 50% because X1 and X2 can be anything in the range?



EDIT: New attempt
$P(X1 ≥ X2) = P(X1 > X2) + P(X1 = X2)$



$P(X1 = X2)$ = $sum_{x}$ $(1-p)^{x-1}p(1-p)^{x-1}p$ = $frac{p}{2-p}$



$P(X1 > X2)$ = $P(X1 < X2)$ and $P(X1 < X2) + P(X1 > X2) + P(X1 = X2) = 1$



Therefore, $P(X1 > X2)$ = $frac{1-P(X1 = X2)}{2}$ = $frac{1-p}{2-p}$
Adding $P(X1 = X2)=frac{p}{2-p}$ to that , I get $P(X1 ≥ X2)$ = $frac{1}{2-p}$



Is this correct?






random-variable geometric-distribution







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Please add the 'self-study' tag.
    $endgroup$
    – StubbornAtom
    5 hours ago






  • 1




    $begingroup$
    Actually because X1 and X2 are discrete variables the equality makes things a bit less obvious.
    $endgroup$
    – usεr11852
    5 hours ago
















3












$begingroup$


Suppose X1 and X2 are independent Geometric p random variables. What is the
probability that X1 ≥ X2?



I am confused about this question because we aren't told anything about X1 and X2 other than they are Geometric. Wouldn't this be 50% because X1 and X2 can be anything in the range?



EDIT: New attempt
$P(X1 ≥ X2) = P(X1 > X2) + P(X1 = X2)$



$P(X1 = X2)$ = $sum_{x}$ $(1-p)^{x-1}p(1-p)^{x-1}p$ = $frac{p}{2-p}$



$P(X1 > X2)$ = $P(X1 < X2)$ and $P(X1 < X2) + P(X1 > X2) + P(X1 = X2) = 1$



Therefore, $P(X1 > X2)$ = $frac{1-P(X1 = X2)}{2}$ = $frac{1-p}{2-p}$
Adding $P(X1 = X2)=frac{p}{2-p}$ to that , I get $P(X1 ≥ X2)$ = $frac{1}{2-p}$



Is this correct?






random-variable geometric-distribution







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Please add the 'self-study' tag.
    $endgroup$
    – StubbornAtom
    5 hours ago






  • 1




    $begingroup$
    Actually because X1 and X2 are discrete variables the equality makes things a bit less obvious.
    $endgroup$
    – usεr11852
    5 hours ago














3












3








3





$begingroup$


Suppose X1 and X2 are independent Geometric p random variables. What is the
probability that X1 ≥ X2?



I am confused about this question because we aren't told anything about X1 and X2 other than they are Geometric. Wouldn't this be 50% because X1 and X2 can be anything in the range?



EDIT: New attempt
$P(X1 ≥ X2) = P(X1 > X2) + P(X1 = X2)$



$P(X1 = X2)$ = $sum_{x}$ $(1-p)^{x-1}p(1-p)^{x-1}p$ = $frac{p}{2-p}$



$P(X1 > X2)$ = $P(X1 < X2)$ and $P(X1 < X2) + P(X1 > X2) + P(X1 = X2) = 1$



Therefore, $P(X1 > X2)$ = $frac{1-P(X1 = X2)}{2}$ = $frac{1-p}{2-p}$
Adding $P(X1 = X2)=frac{p}{2-p}$ to that , I get $P(X1 ≥ X2)$ = $frac{1}{2-p}$



Is this correct?






random-variable geometric-distribution







share|cite|improve this question











$endgroup$




Suppose X1 and X2 are independent Geometric p random variables. What is the
probability that X1 ≥ X2?



I am confused about this question because we aren't told anything about X1 and X2 other than they are Geometric. Wouldn't this be 50% because X1 and X2 can be anything in the range?



EDIT: New attempt
$P(X1 ≥ X2) = P(X1 > X2) + P(X1 = X2)$



$P(X1 = X2)$ = $sum_{x}$ $(1-p)^{x-1}p(1-p)^{x-1}p$ = $frac{p}{2-p}$



$P(X1 > X2)$ = $P(X1 < X2)$ and $P(X1 < X2) + P(X1 > X2) + P(X1 = X2) = 1$



Therefore, $P(X1 > X2)$ = $frac{1-P(X1 = X2)}{2}$ = $frac{1-p}{2-p}$
Adding $P(X1 = X2)=frac{p}{2-p}$ to that , I get $P(X1 ≥ X2)$ = $frac{1}{2-p}$



Is this correct?






random-variable geometric-distribution




random-variable geometric-distribution






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 1 hour ago







Sra

















asked 5 hours ago









SraSra

514




514








  • 1




    $begingroup$
    Please add the 'self-study' tag.
    $endgroup$
    – StubbornAtom
    5 hours ago






  • 1




    $begingroup$
    Actually because X1 and X2 are discrete variables the equality makes things a bit less obvious.
    $endgroup$
    – usεr11852
    5 hours ago














  • 1




    $begingroup$
    Please add the 'self-study' tag.
    $endgroup$
    – StubbornAtom
    5 hours ago






  • 1




    $begingroup$
    Actually because X1 and X2 are discrete variables the equality makes things a bit less obvious.
    $endgroup$
    – usεr11852
    5 hours ago








1




1




$begingroup$
Please add the 'self-study' tag.
$endgroup$
– StubbornAtom
5 hours ago




$begingroup$
Please add the 'self-study' tag.
$endgroup$
– StubbornAtom
5 hours ago




1




1




$begingroup$
Actually because X1 and X2 are discrete variables the equality makes things a bit less obvious.
$endgroup$
– usεr11852
5 hours ago




$begingroup$
Actually because X1 and X2 are discrete variables the equality makes things a bit less obvious.
$endgroup$
– usεr11852
5 hours ago










2 Answers
2






active

oldest

votes


















5












$begingroup$

It can't be $50%$ because $P(X_1=X_2)>0$



One approach:



Consider the three events $P(X_1>X_2), P(X_2>X_1)$ and $P(X_1=X_2)$, which partition the sample space.



There's an obvious connection between the first two. Write an expression for the third and simplify. Hence solve the question.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I edited, my post with my new answer. Could you take a look and see if it's correct?
    $endgroup$
    – Sra
    1 hour ago



















0












$begingroup$

Your answer, following Glen's suggestion, is correct. Another, less elegant, way is just to condition:



begin{align}
Pr{X_1geq X_2} &= sum_{k=0}^infty Pr{X_1geq X_2mid X_2=k} Pr{X_2=k} \ &= sum_{k=0}^infty sum_{ell=k}^infty Pr{X_1=ell}Pr{X_2=k}.
end{align}



This will give you the same $1/(2-p)$, after handling the two geometric series. Glen's way is better.






share|cite|improve this answer









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    2 Answers
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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    5












    $begingroup$

    It can't be $50%$ because $P(X_1=X_2)>0$



    One approach:



    Consider the three events $P(X_1>X_2), P(X_2>X_1)$ and $P(X_1=X_2)$, which partition the sample space.



    There's an obvious connection between the first two. Write an expression for the third and simplify. Hence solve the question.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I edited, my post with my new answer. Could you take a look and see if it's correct?
      $endgroup$
      – Sra
      1 hour ago
















    5












    $begingroup$

    It can't be $50%$ because $P(X_1=X_2)>0$



    One approach:



    Consider the three events $P(X_1>X_2), P(X_2>X_1)$ and $P(X_1=X_2)$, which partition the sample space.



    There's an obvious connection between the first two. Write an expression for the third and simplify. Hence solve the question.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I edited, my post with my new answer. Could you take a look and see if it's correct?
      $endgroup$
      – Sra
      1 hour ago














    5












    5








    5





    $begingroup$

    It can't be $50%$ because $P(X_1=X_2)>0$



    One approach:



    Consider the three events $P(X_1>X_2), P(X_2>X_1)$ and $P(X_1=X_2)$, which partition the sample space.



    There's an obvious connection between the first two. Write an expression for the third and simplify. Hence solve the question.






    share|cite|improve this answer











    $endgroup$



    It can't be $50%$ because $P(X_1=X_2)>0$



    One approach:



    Consider the three events $P(X_1>X_2), P(X_2>X_1)$ and $P(X_1=X_2)$, which partition the sample space.



    There's an obvious connection between the first two. Write an expression for the third and simplify. Hence solve the question.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited 4 hours ago

























    answered 4 hours ago









    Glen_b♦Glen_b

    212k22406754




    212k22406754












    • $begingroup$
      I edited, my post with my new answer. Could you take a look and see if it's correct?
      $endgroup$
      – Sra
      1 hour ago


















    • $begingroup$
      I edited, my post with my new answer. Could you take a look and see if it's correct?
      $endgroup$
      – Sra
      1 hour ago
















    $begingroup$
    I edited, my post with my new answer. Could you take a look and see if it's correct?
    $endgroup$
    – Sra
    1 hour ago




    $begingroup$
    I edited, my post with my new answer. Could you take a look and see if it's correct?
    $endgroup$
    – Sra
    1 hour ago













    0












    $begingroup$

    Your answer, following Glen's suggestion, is correct. Another, less elegant, way is just to condition:



    begin{align}
    Pr{X_1geq X_2} &= sum_{k=0}^infty Pr{X_1geq X_2mid X_2=k} Pr{X_2=k} \ &= sum_{k=0}^infty sum_{ell=k}^infty Pr{X_1=ell}Pr{X_2=k}.
    end{align}



    This will give you the same $1/(2-p)$, after handling the two geometric series. Glen's way is better.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Your answer, following Glen's suggestion, is correct. Another, less elegant, way is just to condition:



      begin{align}
      Pr{X_1geq X_2} &= sum_{k=0}^infty Pr{X_1geq X_2mid X_2=k} Pr{X_2=k} \ &= sum_{k=0}^infty sum_{ell=k}^infty Pr{X_1=ell}Pr{X_2=k}.
      end{align}



      This will give you the same $1/(2-p)$, after handling the two geometric series. Glen's way is better.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Your answer, following Glen's suggestion, is correct. Another, less elegant, way is just to condition:



        begin{align}
        Pr{X_1geq X_2} &= sum_{k=0}^infty Pr{X_1geq X_2mid X_2=k} Pr{X_2=k} \ &= sum_{k=0}^infty sum_{ell=k}^infty Pr{X_1=ell}Pr{X_2=k}.
        end{align}



        This will give you the same $1/(2-p)$, after handling the two geometric series. Glen's way is better.






        share|cite|improve this answer









        $endgroup$



        Your answer, following Glen's suggestion, is correct. Another, less elegant, way is just to condition:



        begin{align}
        Pr{X_1geq X_2} &= sum_{k=0}^infty Pr{X_1geq X_2mid X_2=k} Pr{X_2=k} \ &= sum_{k=0}^infty sum_{ell=k}^infty Pr{X_1=ell}Pr{X_2=k}.
        end{align}



        This will give you the same $1/(2-p)$, after handling the two geometric series. Glen's way is better.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 22 mins ago









        Paulo C. Marques F.Paulo C. Marques F.

        16.9k35397




        16.9k35397






























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