Memoryless property of exponential distribution Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)X1 X2 independent variables exponential distribution - Looking for simpler solutionUsing the Memoryless Property to Explain the Expected Value of the Maximum of iid Exponential RVsMemoryless property and geometric distributionThe Expectation of a function of independent random variablesConditional expectation of an exponential RV, where conditioning is on sum of exponential RVsUnit Measure Axiom for the Gamma DistributionOn the proof that every positive continuous random variable with the memoryless property is exponentially distributedExponential distribution probabilities and memoryless propertyProbability: how to use hint for interarrival time questionDerivation of the kth moment of an exponential distribution
What do you call the holes in a flute?
Windows 10: How to Lock (not sleep) laptop on lid close?
If A makes B more likely then B makes A more likely"
Stars Make Stars
Fishing simulator
Passing functions in C++
Slither Like a Snake
What to do with post with dry rot?
How to add zeros to reach same number of decimal places in tables?
Determine whether f is a function, an injection, a surjection
How to politely respond to generic emails requesting a PhD/job in my lab? Without wasting too much time
Active filter with series inductor and resistor - do these exist?
Is there folklore associating late breastfeeding with low intelligence and/or gullibility?
Autumning in love
New Order #5: where Fibonacci and Beatty meet at Wythoff
What would be Julian Assange's expected punishment, on the current English criminal law?
What are the performance impacts of 'functional' Rust?
Why does this iterative way of solving of equation work?
What is the electric potential inside a point charge?
3 doors, three guards, one stone
What did Darwin mean by 'squib' here?
Geometric mean and geometric standard deviation
How to pour concrete for curved walkway to prevent cracking?
Statistical model of ligand substitution
Memoryless property of exponential distribution
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)X1 X2 independent variables exponential distribution - Looking for simpler solutionUsing the Memoryless Property to Explain the Expected Value of the Maximum of iid Exponential RVsMemoryless property and geometric distributionThe Expectation of a function of independent random variablesConditional expectation of an exponential RV, where conditioning is on sum of exponential RVsUnit Measure Axiom for the Gamma DistributionOn the proof that every positive continuous random variable with the memoryless property is exponentially distributedExponential distribution probabilities and memoryless propertyProbability: how to use hint for interarrival time questionDerivation of the kth moment of an exponential distribution
$begingroup$
In general when $X$ is an exponential random varaible, the memoryless property is stated as
$$mathbbP( X> s+t | X> s ) = mathbbP(X > t).$$
But a direct computation shows if $S$ is also a exponential random variable, then
$$mathbbP( X> S+t | X> S ) = mathbbP(X > t)$$
is true as well. If $Xsim Exp(lambda)$, $Ssim Exp(mu)$, then
$$beginalign
mathbbP( X> S+t, X> S ) &= mathbbP( X> S+t) \
& = int_0^infty int_s+t^infty lambda e^-lambda x mu e^-mu sdxds\
&= int_0^infty mu e^-mu s e^-lambda (s+t) ds\
&= fracmumu+lambda cdot e^-lambda t bigg[int_0^infty (mu+ lambda) e^-(mu+lambda) s dsbigg]\
&= mathbbP(X>S) mathbbP( X>t)
endalign$$
So is there a generalized version of this property? And why don't we introduce it as part of the standard definition.
probability probability-theory
asked Mar 25 at 18:23
XiaoXiao
4,88811636
$endgroup$
add a comment |
$begingroup$
In general when $X$ is an exponential random varaible, the memoryless property is stated as
$$mathbbP( X> s+t | X> s ) = mathbbP(X > t).$$
But a direct computation shows if $S$ is also a exponential random variable, then
$$mathbbP( X> S+t | X> S ) = mathbbP(X > t)$$
is true as well. If $Xsim Exp(lambda)$, $Ssim Exp(mu)$, then
$$beginalign
mathbbP( X> S+t, X> S ) &= mathbbP( X> S+t) \
& = int_0^infty int_s+t^infty lambda e^-lambda x mu e^-mu sdxds\
&= int_0^infty mu e^-mu s e^-lambda (s+t) ds\
&= fracmumu+lambda cdot e^-lambda t bigg[int_0^infty (mu+ lambda) e^-(mu+lambda) s dsbigg]\
&= mathbbP(X>S) mathbbP( X>t)
endalign$$
So is there a generalized version of this property? And why don't we introduce it as part of the standard definition.
probability probability-theory
asked Mar 25 at 18:23
XiaoXiao
4,88811636
$endgroup$
$begingroup$
Are $S$ and $X$ i.i.d ?
$endgroup$
– StubbornAtom
Mar 25 at 19:27
$begingroup$
they are independent exponential r.v., could have different parameters.
$endgroup$
– Xiao
Mar 25 at 20:23
add a comment |
$begingroup$
In general when $X$ is an exponential random varaible, the memoryless property is stated as
$$mathbbP( X> s+t | X> s ) = mathbbP(X > t).$$
But a direct computation shows if $S$ is also a exponential random variable, then
$$mathbbP( X> S+t | X> S ) = mathbbP(X > t)$$
is true as well. If $Xsim Exp(lambda)$, $Ssim Exp(mu)$, then
$$beginalign
mathbbP( X> S+t, X> S ) &= mathbbP( X> S+t) \
& = int_0^infty int_s+t^infty lambda e^-lambda x mu e^-mu sdxds\
&= int_0^infty mu e^-mu s e^-lambda (s+t) ds\
&= fracmumu+lambda cdot e^-lambda t bigg[int_0^infty (mu+ lambda) e^-(mu+lambda) s dsbigg]\
&= mathbbP(X>S) mathbbP( X>t)
endalign$$
So is there a generalized version of this property? And why don't we introduce it as part of the standard definition.
probability probability-theory
asked Mar 25 at 18:23
XiaoXiao
4,88811636
$endgroup$
In general when $X$ is an exponential random varaible, the memoryless property is stated as
$$mathbbP( X> s+t | X> s ) = mathbbP(X > t).$$
But a direct computation shows if $S$ is also a exponential random variable, then
$$mathbbP( X> S+t | X> S ) = mathbbP(X > t)$$
is true as well. If $Xsim Exp(lambda)$, $Ssim Exp(mu)$, then
$$beginalign
mathbbP( X> S+t, X> S ) &= mathbbP( X> S+t) \
& = int_0^infty int_s+t^infty lambda e^-lambda x mu e^-mu sdxds\
&= int_0^infty mu e^-mu s e^-lambda (s+t) ds\
&= fracmumu+lambda cdot e^-lambda t bigg[int_0^infty (mu+ lambda) e^-(mu+lambda) s dsbigg]\
&= mathbbP(X>S) mathbbP( X>t)
endalign$$
So is there a generalized version of this property? And why don't we introduce it as part of the standard definition.
probability probability-theory
probability probability-theory
asked Mar 25 at 18:23
XiaoXiao
4,88811636
asked Mar 25 at 18:23
XiaoXiao
4,88811636
edited Mar 25 at 20:37
Xiao
asked Mar 25 at 18:23
XiaoXiao
4,88811636
asked Mar 25 at 18:23
XiaoXiao
4,88811636
asked Mar 25 at 18:23
XiaoXiao
4,88811636
4,88811636
$begingroup$
Are $S$ and $X$ i.i.d ?
$endgroup$
– StubbornAtom
Mar 25 at 19:27
$begingroup$
they are independent exponential r.v., could have different parameters.
$endgroup$
– Xiao
Mar 25 at 20:23
add a comment |
$begingroup$
Are $S$ and $X$ i.i.d ?
$endgroup$
– StubbornAtom
Mar 25 at 19:27
$begingroup$
they are independent exponential r.v., could have different parameters.
$endgroup$
– Xiao
Mar 25 at 20:23
$begingroup$
Are $S$ and $X$ i.i.d ?
$endgroup$
– StubbornAtom
Mar 25 at 19:27
$begingroup$
Are $S$ and $X$ i.i.d ?
$endgroup$
– StubbornAtom
Mar 25 at 19:27
$begingroup$
they are independent exponential r.v., could have different parameters.
$endgroup$
– Xiao
Mar 25 at 20:23
$begingroup$
they are independent exponential r.v., could have different parameters.
$endgroup$
– Xiao
Mar 25 at 20:23
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
There is nothing particularly interesting going on here. It follows from the memoryless property for $X$, and moreover all that matters about $S$ is that it is a non-negative RV independent of $X$, not that it is exponential: $$ P(X>S+tmid X> S) \= fracP(X>S+t)P(X>S)\=fracint P(X>s+t)f_S(s)dsint P(X>s)f_S(s)ds \=fracint P(X>t)P(X>s)f_S(s)dsint P(X>s)f_S(s)ds \=P(X>t)$$ where in the second to last line we used the memoryless property.
Intuitively, since the conditional distribution is (functionally) independent of the current wait time it really shouldn't matter if the wait time is random or not (as long as it is independently random).
answered Mar 26 at 0:22
spaceisdarkgreenspaceisdarkgreen
34.1k21754
$endgroup$
add a comment |
Your Answer
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
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: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
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
,
noCode: 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%2fmath.stackexchange.com%2fquestions%2f3162152%2fmemoryless-property-of-exponential-distribution%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
There is nothing particularly interesting going on here. It follows from the memoryless property for $X$, and moreover all that matters about $S$ is that it is a non-negative RV independent of $X$, not that it is exponential: $$ P(X>S+tmid X> S) \= fracP(X>S+t)P(X>S)\=fracint P(X>s+t)f_S(s)dsint P(X>s)f_S(s)ds \=fracint P(X>t)P(X>s)f_S(s)dsint P(X>s)f_S(s)ds \=P(X>t)$$ where in the second to last line we used the memoryless property.
Intuitively, since the conditional distribution is (functionally) independent of the current wait time it really shouldn't matter if the wait time is random or not (as long as it is independently random).
answered Mar 26 at 0:22
spaceisdarkgreenspaceisdarkgreen
34.1k21754
$endgroup$
add a comment |
$begingroup$
There is nothing particularly interesting going on here. It follows from the memoryless property for $X$, and moreover all that matters about $S$ is that it is a non-negative RV independent of $X$, not that it is exponential: $$ P(X>S+tmid X> S) \= fracP(X>S+t)P(X>S)\=fracint P(X>s+t)f_S(s)dsint P(X>s)f_S(s)ds \=fracint P(X>t)P(X>s)f_S(s)dsint P(X>s)f_S(s)ds \=P(X>t)$$ where in the second to last line we used the memoryless property.
Intuitively, since the conditional distribution is (functionally) independent of the current wait time it really shouldn't matter if the wait time is random or not (as long as it is independently random).
answered Mar 26 at 0:22
spaceisdarkgreenspaceisdarkgreen
34.1k21754
$endgroup$
add a comment |
$begingroup$
There is nothing particularly interesting going on here. It follows from the memoryless property for $X$, and moreover all that matters about $S$ is that it is a non-negative RV independent of $X$, not that it is exponential: $$ P(X>S+tmid X> S) \= fracP(X>S+t)P(X>S)\=fracint P(X>s+t)f_S(s)dsint P(X>s)f_S(s)ds \=fracint P(X>t)P(X>s)f_S(s)dsint P(X>s)f_S(s)ds \=P(X>t)$$ where in the second to last line we used the memoryless property.
Intuitively, since the conditional distribution is (functionally) independent of the current wait time it really shouldn't matter if the wait time is random or not (as long as it is independently random).
answered Mar 26 at 0:22
spaceisdarkgreenspaceisdarkgreen
34.1k21754
$endgroup$
There is nothing particularly interesting going on here. It follows from the memoryless property for $X$, and moreover all that matters about $S$ is that it is a non-negative RV independent of $X$, not that it is exponential: $$ P(X>S+tmid X> S) \= fracP(X>S+t)P(X>S)\=fracint P(X>s+t)f_S(s)dsint P(X>s)f_S(s)ds \=fracint P(X>t)P(X>s)f_S(s)dsint P(X>s)f_S(s)ds \=P(X>t)$$ where in the second to last line we used the memoryless property.
Intuitively, since the conditional distribution is (functionally) independent of the current wait time it really shouldn't matter if the wait time is random or not (as long as it is independently random).
answered Mar 26 at 0:22
spaceisdarkgreenspaceisdarkgreen
34.1k21754
answered Mar 26 at 0:22
spaceisdarkgreenspaceisdarkgreen
34.1k21754
answered Mar 26 at 0:22
spaceisdarkgreenspaceisdarkgreen
34.1k21754
answered Mar 26 at 0:22
spaceisdarkgreenspaceisdarkgreen
34.1k21754
34.1k21754
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- 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%2fmath.stackexchange.com%2fquestions%2f3162152%2fmemoryless-property-of-exponential-distribution%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
$begingroup$
Are $S$ and $X$ i.i.d ?
$endgroup$
– StubbornAtom
Mar 25 at 19:27
$begingroup$
they are independent exponential r.v., could have different parameters.
$endgroup$
– Xiao
Mar 25 at 20:23