Probability of $X_1 geq X_2$
Clash Royale CLAN TAG#URR8PPP
$begingroup$
Suppose $X_1$ and $X_2$ are independent geometric random variables with parameter $p$. What is the probability that $X_1 geq X_2$?
I am confused about this question because we aren't told anything about $X_1$ and $X_2$ other than they are geometric. Wouldn't this be $50%$ because $X_1$ and $X_2$ can be anything in the range?
EDIT: New attempt
$P(X1 ≥ X2) = P(X1 > X2) + P(X1 = X2)$
$P(X1 = X2)$ = $sum_x$ $(1-p)^x-1p(1-p)^x-1p$ = $fracp2-p$
$P(X1 > X2)$ = $P(X1 < X2)$ and $P(X1 < X2) + P(X1 > X2) + P(X1 = X2) = 1$
Therefore, $P(X1 > X2)$ = $frac1-P(X1 = X2)2$ = $frac1-p2-p$
Adding $P(X1 = X2)=fracp2-p$ to that , I get $P(X1 ≥ X2)$ = $frac12-p$
Is this correct?
self-study random-variable geometric-distribution
edited Feb 24 at 16:51
StatsStudent
6,04332044
asked Feb 23 at 23:15
SraSra
944
$endgroup$
add a comment |
$begingroup$
Suppose $X_1$ and $X_2$ are independent geometric random variables with parameter $p$. What is the probability that $X_1 geq X_2$?
I am confused about this question because we aren't told anything about $X_1$ and $X_2$ other than they are geometric. Wouldn't this be $50%$ because $X_1$ and $X_2$ can be anything in the range?
EDIT: New attempt
$P(X1 ≥ X2) = P(X1 > X2) + P(X1 = X2)$
$P(X1 = X2)$ = $sum_x$ $(1-p)^x-1p(1-p)^x-1p$ = $fracp2-p$
$P(X1 > X2)$ = $P(X1 < X2)$ and $P(X1 < X2) + P(X1 > X2) + P(X1 = X2) = 1$
Therefore, $P(X1 > X2)$ = $frac1-P(X1 = X2)2$ = $frac1-p2-p$
Adding $P(X1 = X2)=fracp2-p$ to that , I get $P(X1 ≥ X2)$ = $frac12-p$
Is this correct?
self-study random-variable geometric-distribution
edited Feb 24 at 16:51
StatsStudent
6,04332044
asked Feb 23 at 23:15
SraSra
944
$endgroup$
3
$begingroup$
Please add the 'self-study' tag.
$endgroup$
– StubbornAtom
Feb 23 at 23:21
1
$begingroup$
Actually becauseX1andX2are discrete variables the equality makes things a bit less obvious.
$endgroup$
– usεr11852
Feb 23 at 23:21
add a comment |
$begingroup$
Suppose $X_1$ and $X_2$ are independent geometric random variables with parameter $p$. What is the probability that $X_1 geq X_2$?
I am confused about this question because we aren't told anything about $X_1$ and $X_2$ other than they are geometric. Wouldn't this be $50%$ because $X_1$ and $X_2$ can be anything in the range?
EDIT: New attempt
$P(X1 ≥ X2) = P(X1 > X2) + P(X1 = X2)$
$P(X1 = X2)$ = $sum_x$ $(1-p)^x-1p(1-p)^x-1p$ = $fracp2-p$
$P(X1 > X2)$ = $P(X1 < X2)$ and $P(X1 < X2) + P(X1 > X2) + P(X1 = X2) = 1$
Therefore, $P(X1 > X2)$ = $frac1-P(X1 = X2)2$ = $frac1-p2-p$
Adding $P(X1 = X2)=fracp2-p$ to that , I get $P(X1 ≥ X2)$ = $frac12-p$
Is this correct?
self-study random-variable geometric-distribution
edited Feb 24 at 16:51
StatsStudent
6,04332044
asked Feb 23 at 23:15
SraSra
944
$endgroup$
Suppose $X_1$ and $X_2$ are independent geometric random variables with parameter $p$. What is the probability that $X_1 geq X_2$?
I am confused about this question because we aren't told anything about $X_1$ and $X_2$ other than they are geometric. Wouldn't this be $50%$ because $X_1$ and $X_2$ can be anything in the range?
EDIT: New attempt
$P(X1 ≥ X2) = P(X1 > X2) + P(X1 = X2)$
$P(X1 = X2)$ = $sum_x$ $(1-p)^x-1p(1-p)^x-1p$ = $fracp2-p$
$P(X1 > X2)$ = $P(X1 < X2)$ and $P(X1 < X2) + P(X1 > X2) + P(X1 = X2) = 1$
Therefore, $P(X1 > X2)$ = $frac1-P(X1 = X2)2$ = $frac1-p2-p$
Adding $P(X1 = X2)=fracp2-p$ to that , I get $P(X1 ≥ X2)$ = $frac12-p$
Is this correct?
self-study random-variable geometric-distribution
self-study random-variable geometric-distribution
edited Feb 24 at 16:51
StatsStudent
6,04332044
asked Feb 23 at 23:15
SraSra
944
edited Feb 24 at 16:51
StatsStudent
6,04332044
asked Feb 23 at 23:15
SraSra
944
edited Feb 24 at 16:51
StatsStudent
6,04332044
edited Feb 24 at 16:51
StatsStudent
6,04332044
edited Feb 24 at 16:51
StatsStudent
6,04332044
6,04332044
asked Feb 23 at 23:15
SraSra
944
asked Feb 23 at 23:15
SraSra
944
asked Feb 23 at 23:15
SraSra
944
944
3
$begingroup$
Please add the 'self-study' tag.
$endgroup$
– StubbornAtom
Feb 23 at 23:21
1
$begingroup$
Actually becauseX1andX2are discrete variables the equality makes things a bit less obvious.
$endgroup$
– usεr11852
Feb 23 at 23:21
add a comment |
3
$begingroup$
Please add the 'self-study' tag.
$endgroup$
– StubbornAtom
Feb 23 at 23:21
1
$begingroup$
Actually becauseX1andX2are discrete variables the equality makes things a bit less obvious.
$endgroup$
– usεr11852
Feb 23 at 23:21
3
3
$begingroup$
Please add the 'self-study' tag.
$endgroup$
– StubbornAtom
Feb 23 at 23:21
$begingroup$
Please add the 'self-study' tag.
$endgroup$
– StubbornAtom
Feb 23 at 23:21
1
1
$begingroup$
Actually because
X1 and X2 are discrete variables the equality makes things a bit less obvious.$endgroup$
– usεr11852
Feb 23 at 23:21
$begingroup$
Actually because
X1 and X2 are discrete variables the equality makes things a bit less obvious.$endgroup$
– usεr11852
Feb 23 at 23:21
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 Feb 23 at 23:55
Glen_b♦Glen_b
214k23415765
$endgroup$
$begingroup$
I edited, my post with my new answer. Could you take a look and see if it's correct?
$endgroup$
– Sra
Feb 24 at 3:02
1
$begingroup$
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♦
Feb 24 at 4:26
add a comment |
$begingroup$
Your answer, following Glen's suggestion, is correct. Another, less elegant, way is just to condition:
beginalign
PrX_1geq X_2 &= sum_k=0^infty PrX_1geq X_2mid X_2=k PrX_2=k \ &= sum_k=0^infty sum_ell=k^infty PrX_1=ellPrX_2=k.
endalign
This will give you the same $1/(2-p)$, after handling the two geometric series. Glen's way is better.
answered Feb 24 at 4:01
ZenZen
17.3k35498
$endgroup$
4
$begingroup$
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
Feb 24 at 4:28
3
$begingroup$
@probabilityislogic I share your enthusiasm for derivations from "first principles." However, to a modern mathematician, looking for and exploiting symmetry is even more fundamental than the first principles (definitions) to which you refer: we might call it a metaprinciple of mathematics. It's much more than a mere "trick."
$endgroup$
– whuber♦
Feb 24 at 17:13
add a comment |
Your Answer
StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "65"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstats.stackexchange.com%2fquestions%2f394042%2fprobability-of-x-1-geq-x-2%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
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 Feb 23 at 23:55
Glen_b♦Glen_b
214k23415765
$endgroup$
$begingroup$
I edited, my post with my new answer. Could you take a look and see if it's correct?
$endgroup$
– Sra
Feb 24 at 3:02
1
$begingroup$
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♦
Feb 24 at 4:26
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 Feb 23 at 23:55
Glen_b♦Glen_b
214k23415765
$endgroup$
$begingroup$
I edited, my post with my new answer. Could you take a look and see if it's correct?
$endgroup$
– Sra
Feb 24 at 3:02
1
$begingroup$
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♦
Feb 24 at 4:26
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 Feb 23 at 23:55
Glen_b♦Glen_b
214k23415765
$endgroup$
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 Feb 23 at 23:55
Glen_b♦Glen_b
214k23415765
edited Feb 24 at 0:03
answered Feb 23 at 23:55
Glen_b♦Glen_b
214k23415765
answered Feb 23 at 23:55
Glen_b♦Glen_b
214k23415765
answered Feb 23 at 23:55
Glen_b♦Glen_b
214k23415765
214k23415765
$begingroup$
I edited, my post with my new answer. Could you take a look and see if it's correct?
$endgroup$
– Sra
Feb 24 at 3:02
1
$begingroup$
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♦
Feb 24 at 4:26
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
Feb 24 at 3:02
1
$begingroup$
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♦
Feb 24 at 4:26
$begingroup$
I edited, my post with my new answer. Could you take a look and see if it's correct?
$endgroup$
– Sra
Feb 24 at 3:02
$begingroup$
I edited, my post with my new answer. Could you take a look and see if it's correct?
$endgroup$
– Sra
Feb 24 at 3:02
1
1
$begingroup$
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♦
Feb 24 at 4:26
$begingroup$
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♦
Feb 24 at 4:26
add a comment |
$begingroup$
Your answer, following Glen's suggestion, is correct. Another, less elegant, way is just to condition:
beginalign
PrX_1geq X_2 &= sum_k=0^infty PrX_1geq X_2mid X_2=k PrX_2=k \ &= sum_k=0^infty sum_ell=k^infty PrX_1=ellPrX_2=k.
endalign
This will give you the same $1/(2-p)$, after handling the two geometric series. Glen's way is better.
answered Feb 24 at 4:01
ZenZen
17.3k35498
$endgroup$
4
$begingroup$
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
Feb 24 at 4:28
3
$begingroup$
@probabilityislogic I share your enthusiasm for derivations from "first principles." However, to a modern mathematician, looking for and exploiting symmetry is even more fundamental than the first principles (definitions) to which you refer: we might call it a metaprinciple of mathematics. It's much more than a mere "trick."
$endgroup$
– whuber♦
Feb 24 at 17:13
add a comment |
$begingroup$
Your answer, following Glen's suggestion, is correct. Another, less elegant, way is just to condition:
beginalign
PrX_1geq X_2 &= sum_k=0^infty PrX_1geq X_2mid X_2=k PrX_2=k \ &= sum_k=0^infty sum_ell=k^infty PrX_1=ellPrX_2=k.
endalign
This will give you the same $1/(2-p)$, after handling the two geometric series. Glen's way is better.
answered Feb 24 at 4:01
ZenZen
17.3k35498
$endgroup$
4
$begingroup$
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
Feb 24 at 4:28
3
$begingroup$
@probabilityislogic I share your enthusiasm for derivations from "first principles." However, to a modern mathematician, looking for and exploiting symmetry is even more fundamental than the first principles (definitions) to which you refer: we might call it a metaprinciple of mathematics. It's much more than a mere "trick."
$endgroup$
– whuber♦
Feb 24 at 17:13
add a comment |
$begingroup$
Your answer, following Glen's suggestion, is correct. Another, less elegant, way is just to condition:
beginalign
PrX_1geq X_2 &= sum_k=0^infty PrX_1geq X_2mid X_2=k PrX_2=k \ &= sum_k=0^infty sum_ell=k^infty PrX_1=ellPrX_2=k.
endalign
This will give you the same $1/(2-p)$, after handling the two geometric series. Glen's way is better.
answered Feb 24 at 4:01
ZenZen
17.3k35498
$endgroup$
Your answer, following Glen's suggestion, is correct. Another, less elegant, way is just to condition:
beginalign
PrX_1geq X_2 &= sum_k=0^infty PrX_1geq X_2mid X_2=k PrX_2=k \ &= sum_k=0^infty sum_ell=k^infty PrX_1=ellPrX_2=k.
endalign
This will give you the same $1/(2-p)$, after handling the two geometric series. Glen's way is better.
answered Feb 24 at 4:01
ZenZen
17.3k35498
answered Feb 24 at 4:01
ZenZen
17.3k35498
answered Feb 24 at 4:01
ZenZen
17.3k35498
answered Feb 24 at 4:01
ZenZen
17.3k35498
17.3k35498
4
$begingroup$
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
Feb 24 at 4:28
3
$begingroup$
@probabilityislogic I share your enthusiasm for derivations from "first principles." However, to a modern mathematician, looking for and exploiting symmetry is even more fundamental than the first principles (definitions) to which you refer: we might call it a metaprinciple of mathematics. It's much more than a mere "trick."
$endgroup$
– whuber♦
Feb 24 at 17:13
add a comment |
4
$begingroup$
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
Feb 24 at 4:28
3
$begingroup$
@probabilityislogic I share your enthusiasm for derivations from "first principles." However, to a modern mathematician, looking for and exploiting symmetry is even more fundamental than the first principles (definitions) to which you refer: we might call it a metaprinciple of mathematics. It's much more than a mere "trick."
$endgroup$
– whuber♦
Feb 24 at 17:13
4
4
$begingroup$
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
Feb 24 at 4:28
$begingroup$
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
Feb 24 at 4:28
3
3
$begingroup$
@probabilityislogic I share your enthusiasm for derivations from "first principles." However, to a modern mathematician, looking for and exploiting symmetry is even more fundamental than the first principles (definitions) to which you refer: we might call it a metaprinciple of mathematics. It's much more than a mere "trick."
$endgroup$
– whuber♦
Feb 24 at 17:13
$begingroup$
@probabilityislogic I share your enthusiasm for derivations from "first principles." However, to a modern mathematician, looking for and exploiting symmetry is even more fundamental than the first principles (definitions) to which you refer: we might call it a metaprinciple of mathematics. It's much more than a mere "trick."
$endgroup$
– whuber♦
Feb 24 at 17:13
add a comment |
Thanks for contributing an answer to Cross Validated!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstats.stackexchange.com%2fquestions%2f394042%2fprobability-of-x-1-geq-x-2%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
3
$begingroup$
Please add the 'self-study' tag.
$endgroup$
– StubbornAtom
Feb 23 at 23:21
1
$begingroup$
Actually because
X1andX2are discrete variables the equality makes things a bit less obvious.$endgroup$
– usεr11852
Feb 23 at 23:21