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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
$begingroup$
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
asked 5 hours ago
SraSra
464
$endgroup$
add a comment |
$begingroup$
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
asked 5 hours ago
SraSra
464
$endgroup$
1
$begingroup$
Please add the 'self-study' tag.
$endgroup$
– StubbornAtom
4 hours ago
1
$begingroup$
Actually becauseX1andX2are discrete variables the equality makes things a bit less obvious.
$endgroup$
– usεr11852
4 hours ago
add a comment |
$begingroup$
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
asked 5 hours ago
SraSra
464
$endgroup$
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
asked 5 hours ago
SraSra
464
asked 5 hours ago
SraSra
464
edited 1 hour ago
Sra
asked 5 hours ago
SraSra
464
asked 5 hours ago
SraSra
464
asked 5 hours ago
SraSra
464
464
1
$begingroup$
Please add the 'self-study' tag.
$endgroup$
– StubbornAtom
4 hours ago
1
$begingroup$
Actually becauseX1andX2are discrete variables the equality makes things a bit less obvious.
$endgroup$
– usεr11852
4 hours ago
add a comment |
1
$begingroup$
Please add the 'self-study' tag.
$endgroup$
– StubbornAtom
4 hours ago
1
$begingroup$
Actually becauseX1andX2are 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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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.
answered 4 hours ago
Glen_b♦Glen_b
212k22406754
$endgroup$
$begingroup$
I edited, my post with my new answer. Could you take a look and see if it's correct?
$endgroup$
– Sra
1 hour ago
add a comment |
$begingroup$
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.
answered 19 mins ago
Paulo C. Marques F.Paulo C. Marques F.
16.9k35397
$endgroup$
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
answered 4 hours ago
Glen_b♦Glen_b
212k22406754
$endgroup$
$begingroup$
I edited, my post with my new answer. Could you take a look and see if it's correct?
$endgroup$
– Sra
1 hour ago
add a comment |
$begingroup$
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.
answered 4 hours ago
Glen_b♦Glen_b
212k22406754
$endgroup$
$begingroup$
I edited, my post with my new answer. Could you take a look and see if it's correct?
$endgroup$
– Sra
1 hour ago
add a comment |
$begingroup$
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.
answered 4 hours ago
Glen_b♦Glen_b
212k22406754
$endgroup$
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.
answered 4 hours ago
Glen_b♦Glen_b
212k22406754
edited 4 hours ago
answered 4 hours ago
Glen_b♦Glen_b
212k22406754
answered 4 hours ago
Glen_b♦Glen_b
212k22406754
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
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.
answered 19 mins ago
Paulo C. Marques F.Paulo C. Marques F.
16.9k35397
$endgroup$
add a comment |
$begingroup$
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.
answered 19 mins ago
Paulo C. Marques F.Paulo C. Marques F.
16.9k35397
$endgroup$
add a comment |
$begingroup$
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.
answered 19 mins ago
Paulo C. Marques F.Paulo C. Marques F.
16.9k35397
$endgroup$
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.
answered 19 mins ago
Paulo C. Marques F.Paulo C. Marques F.
16.9k35397
answered 19 mins ago
Paulo C. Marques F.Paulo C. Marques F.
16.9k35397
answered 19 mins ago
Paulo C. Marques F.Paulo C. Marques F.
16.9k35397
answered 19 mins ago
Paulo C. Marques F.Paulo C. Marques F.
16.9k35397
16.9k35397
add a comment |
add a comment |
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1
$begingroup$
Please add the 'self-study' tag.
$endgroup$
– StubbornAtom
4 hours ago
1
$begingroup$
Actually because
X1andX2are discrete variables the equality makes things a bit less obvious.$endgroup$
– usεr11852
4 hours ago