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Probability X1 ≥ X2


In convergence in probability or a.s. convergence w.r.t which measure is the probability?Comparison of waiting times to geometric distributionShow that for a Geometric distribution, the probability generating function is given by $frac{ps}{1-qs}$, $q=1-p$Different ways of generating Geometric distributionsHow to calculate the sum or difference of two probability generating functions?How to check if functions of i.i.d random variables are dependent or independent?Calculating probability of floor'ed log-normal distributionConcepts of Probability all messed upGeometric distribution with random, varying success probabilityWhat can we say about $N_{i}$ where $N=N_{1}+cdots+N_{m}$, $Nthicksim Geom(frac{1-p}{p})$ and conditional distribution of $N_{j}$ is binomial













2












$begingroup$


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
    4 hours ago






  • 1




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
















2












$begingroup$


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
    4 hours ago






  • 1




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














2












2








2





$begingroup$


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

464




464








  • 1




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






  • 1




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














  • 1




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






  • 1




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








1




1




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




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




1




1




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




$begingroup$
Actually because X1 and X2 are discrete variables the equality makes things a bit less obvious.
$endgroup$
– usεr11852
4 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









$endgroup$













    Your Answer





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    2 Answers
    2






    active

    oldest

    votes








    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 16 mins ago









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

        16.9k35397




        16.9k35397






























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