Finding the probabilty $P(Y>S| Y>t)$ for an exponential probabilty distribution. [on hold]

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My question is as follows and also my attempted solution.



The R.V Y has an exp distribution with PDF



$$f(y)=left{beginmatrix lambda e^-lambda y,;;;; y>0\
0,;;; textotherwise endmatrixright.$$
Showing workings, find $P(Y>s| Y>t)$, for some $sgeq t$
and derive an expression for the conditional pdf of $Y$, conditional on $Yleq200$.



I want to use the memorylessness property here, and doing that for a start gives me $P(Y>t-s)$ which is by definition $0$? Is that the solution?




probability




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put on hold as unclear what you're asking by Did, max_zorn, Michael Rozenberg, amWhy, Shailesh yesterday


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.










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    Not $P(Y>t-s)$ but $P(Y>s-t)$. Numerically speaking, memorylesness let's you say that $P(X>50|Xgeq 30) = P(X>20)$ for example
    – Davide Morgante
    Jul 28 at 13:46














up vote
-1
down vote

favorite












My question is as follows and also my attempted solution.



The R.V Y has an exp distribution with PDF



$$f(y)=left{beginmatrix lambda e^-lambda y,;;;; y>0\
0,;;; textotherwise endmatrixright.$$
Showing workings, find $P(Y>s| Y>t)$, for some $sgeq t$
and derive an expression for the conditional pdf of $Y$, conditional on $Yleq200$.



I want to use the memorylessness property here, and doing that for a start gives me $P(Y>t-s)$ which is by definition $0$? Is that the solution?




probability




share|cite|improve this question













put on hold as unclear what you're asking by Did, max_zorn, Michael Rozenberg, amWhy, Shailesh yesterday


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.










  • 1




    Not $P(Y>t-s)$ but $P(Y>s-t)$. Numerically speaking, memorylesness let's you say that $P(X>50|Xgeq 30) = P(X>20)$ for example
    – Davide Morgante
    Jul 28 at 13:46












up vote
-1
down vote

favorite









up vote
-1
down vote

favorite











My question is as follows and also my attempted solution.



The R.V Y has an exp distribution with PDF



$$f(y)=left{beginmatrix lambda e^-lambda y,;;;; y>0\
0,;;; textotherwise endmatrixright.$$
Showing workings, find $P(Y>s| Y>t)$, for some $sgeq t$
and derive an expression for the conditional pdf of $Y$, conditional on $Yleq200$.



I want to use the memorylessness property here, and doing that for a start gives me $P(Y>t-s)$ which is by definition $0$? Is that the solution?




probability




share|cite|improve this question













My question is as follows and also my attempted solution.



The R.V Y has an exp distribution with PDF



$$f(y)=left{beginmatrix lambda e^-lambda y,;;;; y>0\
0,;;; textotherwise endmatrixright.$$
Showing workings, find $P(Y>s| Y>t)$, for some $sgeq t$
and derive an expression for the conditional pdf of $Y$, conditional on $Yleq200$.



I want to use the memorylessness property here, and doing that for a start gives me $P(Y>t-s)$ which is by definition $0$? Is that the solution?





probability





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 28 at 13:55









Davide Morgante

1,724220




1,724220









asked Jul 28 at 13:26









Robbie Meaney

75




75




put on hold as unclear what you're asking by Did, max_zorn, Michael Rozenberg, amWhy, Shailesh yesterday


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.






put on hold as unclear what you're asking by Did, max_zorn, Michael Rozenberg, amWhy, Shailesh yesterday


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.









  • 1




    Not $P(Y>t-s)$ but $P(Y>s-t)$. Numerically speaking, memorylesness let's you say that $P(X>50|Xgeq 30) = P(X>20)$ for example
    – Davide Morgante
    Jul 28 at 13:46












  • 1




    Not $P(Y>t-s)$ but $P(Y>s-t)$. Numerically speaking, memorylesness let's you say that $P(X>50|Xgeq 30) = P(X>20)$ for example
    – Davide Morgante
    Jul 28 at 13:46







1




1




Not $P(Y>t-s)$ but $P(Y>s-t)$. Numerically speaking, memorylesness let's you say that $P(X>50|Xgeq 30) = P(X>20)$ for example
– Davide Morgante
Jul 28 at 13:46




Not $P(Y>t-s)$ but $P(Y>s-t)$. Numerically speaking, memorylesness let's you say that $P(X>50|Xgeq 30) = P(X>20)$ for example
– Davide Morgante
Jul 28 at 13:46










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Your question is confusing. Do you want the distribution of $Y$ under the condition $(tlt Ylt 200)$? It looks like the CDF would be $F(y)=frace^-lambda t-e^-lambda ye^-lambda t-e^-lambda 200$. Also $F(y)=1. ygt 200$.






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    1 Answer
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    up vote
    0
    down vote













    Your question is confusing. Do you want the distribution of $Y$ under the condition $(tlt Ylt 200)$? It looks like the CDF would be $F(y)=frace^-lambda t-e^-lambda ye^-lambda t-e^-lambda 200$. Also $F(y)=1. ygt 200$.






    share|cite|improve this answer

























      up vote
      0
      down vote













      Your question is confusing. Do you want the distribution of $Y$ under the condition $(tlt Ylt 200)$? It looks like the CDF would be $F(y)=frace^-lambda t-e^-lambda ye^-lambda t-e^-lambda 200$. Also $F(y)=1. ygt 200$.






      share|cite|improve this answer























        up vote
        0
        down vote










        up vote
        0
        down vote









        Your question is confusing. Do you want the distribution of $Y$ under the condition $(tlt Ylt 200)$? It looks like the CDF would be $F(y)=frace^-lambda t-e^-lambda ye^-lambda t-e^-lambda 200$. Also $F(y)=1. ygt 200$.






        share|cite|improve this answer













        Your question is confusing. Do you want the distribution of $Y$ under the condition $(tlt Ylt 200)$? It looks like the CDF would be $F(y)=frace^-lambda t-e^-lambda ye^-lambda t-e^-lambda 200$. Also $F(y)=1. ygt 200$.







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered Jul 28 at 15:45









        herb steinberg

        93529




        93529












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