Probability X1 ≥ X2












4












$begingroup$


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
















4












$begingroup$


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














4












4








4





$begingroup$


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

584




584








  • 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


















6












$begingroup$

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










  • $begingroup$
    Yes, your answers look correct. An alternative method (using a similar idea) would be to note that $P(X_1geq X_2)=frac12 + frac12 P(X_1=X_2)$ (again, exploiting the symmetry/exchangeability of $X_1$ and $X_2$).
    $endgroup$
    – Glen_b♦
    8 mins ago





















2












$begingroup$

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









$endgroup$













  • $begingroup$
    note - your way is better for applying to new problems I think. Because it is based on first principles. The trick/intuiton from glen_b's answer usually comes after the problem has been solved your way
    $endgroup$
    – probabilityislogic
    6 mins ago













Your Answer





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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









6












$begingroup$

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










  • $begingroup$
    Yes, your answers look correct. An alternative method (using a similar idea) would be to note that $P(X_1geq X_2)=frac12 + frac12 P(X_1=X_2)$ (again, exploiting the symmetry/exchangeability of $X_1$ and $X_2$).
    $endgroup$
    – Glen_b♦
    8 mins ago


















6












$begingroup$

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










  • $begingroup$
    Yes, your answers look correct. An alternative method (using a similar idea) would be to note that $P(X_1geq X_2)=frac12 + frac12 P(X_1=X_2)$ (again, exploiting the symmetry/exchangeability of $X_1$ and $X_2$).
    $endgroup$
    – Glen_b♦
    8 mins ago
















6












6








6





$begingroup$

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










  • $begingroup$
    Yes, your answers look correct. An alternative method (using a similar idea) would be to note that $P(X_1geq X_2)=frac12 + frac12 P(X_1=X_2)$ (again, exploiting the symmetry/exchangeability of $X_1$ and $X_2$).
    $endgroup$
    – Glen_b♦
    8 mins 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










  • $begingroup$
    Yes, your answers look correct. An alternative method (using a similar idea) would be to note that $P(X_1geq X_2)=frac12 + frac12 P(X_1=X_2)$ (again, exploiting the symmetry/exchangeability of $X_1$ and $X_2$).
    $endgroup$
    – Glen_b♦
    8 mins 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












$begingroup$
Yes, your answers look correct. An alternative method (using a similar idea) would be to note that $P(X_1geq X_2)=frac12 + frac12 P(X_1=X_2)$ (again, exploiting the symmetry/exchangeability of $X_1$ and $X_2$).
$endgroup$
– Glen_b♦
8 mins ago






$begingroup$
Yes, your answers look correct. An alternative method (using a similar idea) would be to note that $P(X_1geq X_2)=frac12 + frac12 P(X_1=X_2)$ (again, exploiting the symmetry/exchangeability of $X_1$ and $X_2$).
$endgroup$
– Glen_b♦
8 mins ago















2












$begingroup$

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









$endgroup$













  • $begingroup$
    note - your way is better for applying to new problems I think. Because it is based on first principles. The trick/intuiton from glen_b's answer usually comes after the problem has been solved your way
    $endgroup$
    – probabilityislogic
    6 mins ago


















2












$begingroup$

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









$endgroup$













  • $begingroup$
    note - your way is better for applying to new problems I think. Because it is based on first principles. The trick/intuiton from glen_b's answer usually comes after the problem has been solved your way
    $endgroup$
    – probabilityislogic
    6 mins ago
















2












2








2





$begingroup$

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 33 mins ago









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

16.9k35397




16.9k35397












  • $begingroup$
    note - your way is better for applying to new problems I think. Because it is based on first principles. The trick/intuiton from glen_b's answer usually comes after the problem has been solved your way
    $endgroup$
    – probabilityislogic
    6 mins ago




















  • $begingroup$
    note - your way is better for applying to new problems I think. Because it is based on first principles. The trick/intuiton from glen_b's answer usually comes after the problem has been solved your way
    $endgroup$
    – probabilityislogic
    6 mins ago


















$begingroup$
note - your way is better for applying to new problems I think. Because it is based on first principles. The trick/intuiton from glen_b's answer usually comes after the problem has been solved your way
$endgroup$
– probabilityislogic
6 mins ago






$begingroup$
note - your way is better for applying to new problems I think. Because it is based on first principles. The trick/intuiton from glen_b's answer usually comes after the problem has been solved your way
$endgroup$
– probabilityislogic
6 mins ago




















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