Probability X1 ≥ X2
$begingroup$
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 4 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 4 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 4 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 4 hours ago
SraSra
464
asked 4 hours ago
SraSra
464
edited 56 mins ago
Sra
asked 4 hours ago
SraSra
464
asked 4 hours ago
SraSra
464
asked 4 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 |
1 Answer
1
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
56 mins ago
add a comment |
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$begingroup$
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
56 mins 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
56 mins 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 3 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
56 mins 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
56 mins ago
$begingroup$
I edited, my post with my new answer. Could you take a look and see if it's correct?
$endgroup$
– Sra
56 mins ago
$begingroup$
I edited, my post with my new answer. Could you take a look and see if it's correct?
$endgroup$
– Sra
56 mins ago
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