Probability that lightbulb stops working in odd yearProbability about random variables with exponential distributionMaximum likelihood estimator of minimum function with exponential RV and a random numberProbability: Hazard Rate functionFind joint density function of X and X+Y (exponential distribution)CDF of two exponential variablesProbability exercise (exponential variable)Inconsistent interpretations for the second parameter of the Gamma distributionExponential distribution probability independent eventsProbability of computer server being online without crashingShow that $P(N geq n) = (1-e^-lambda)^n/ lambda^n $

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Probability that lightbulb stops working in odd year


Probability about random variables with exponential distributionMaximum likelihood estimator of minimum function with exponential RV and a random numberProbability: Hazard Rate functionFind joint density function of X and X+Y (exponential distribution)CDF of two exponential variablesProbability exercise (exponential variable)Inconsistent interpretations for the second parameter of the Gamma distributionExponential distribution probability independent eventsProbability of computer server being online without crashingShow that $P(N geq n) = (1-e^-lambda)^n/ lambda^n $













1












$begingroup$


I have a lightbulb that has an exponential lifetime distribution with mean $mu$ months. So if I construct a pdf $f(x)$ and cdf $F(x)$ with parameter $lambda$, and since $E[X] = frac1lambda$,



beginalign
f(x) &= frac1mue^-frac1mux \
F(x) &= 1-e^-frac1mux
endalign



Year $Y$ can be odd if it's the $1$st year ($12$ months), $3$rd year ($36$ months) and so on.



I'm having troubles constructing the $P(Y= mathrmodd)$ model here, as I don't know how to put together the infinite odd years described above.






probability exponential-distribution







share|cite|improve this question











$endgroup$







  • 2




    $begingroup$
    If $X$ is the random month in which the lightbulb kaputs, and $Y$ is the year of that happening, then $Y$ is odd whenever $X$ is between $1$ and $12$, or $25$ and $36$, or $49$ and $60$, and so on.
    $endgroup$
    – Rócherz
    Mar 15 at 1:21















1












$begingroup$


I have a lightbulb that has an exponential lifetime distribution with mean $mu$ months. So if I construct a pdf $f(x)$ and cdf $F(x)$ with parameter $lambda$, and since $E[X] = frac1lambda$,



beginalign
f(x) &= frac1mue^-frac1mux \
F(x) &= 1-e^-frac1mux
endalign



Year $Y$ can be odd if it's the $1$st year ($12$ months), $3$rd year ($36$ months) and so on.



I'm having troubles constructing the $P(Y= mathrmodd)$ model here, as I don't know how to put together the infinite odd years described above.






probability exponential-distribution







share|cite|improve this question











$endgroup$







  • 2




    $begingroup$
    If $X$ is the random month in which the lightbulb kaputs, and $Y$ is the year of that happening, then $Y$ is odd whenever $X$ is between $1$ and $12$, or $25$ and $36$, or $49$ and $60$, and so on.
    $endgroup$
    – Rócherz
    Mar 15 at 1:21













1












1








1





$begingroup$


I have a lightbulb that has an exponential lifetime distribution with mean $mu$ months. So if I construct a pdf $f(x)$ and cdf $F(x)$ with parameter $lambda$, and since $E[X] = frac1lambda$,



beginalign
f(x) &= frac1mue^-frac1mux \
F(x) &= 1-e^-frac1mux
endalign



Year $Y$ can be odd if it's the $1$st year ($12$ months), $3$rd year ($36$ months) and so on.



I'm having troubles constructing the $P(Y= mathrmodd)$ model here, as I don't know how to put together the infinite odd years described above.






probability exponential-distribution







share|cite|improve this question











$endgroup$




I have a lightbulb that has an exponential lifetime distribution with mean $mu$ months. So if I construct a pdf $f(x)$ and cdf $F(x)$ with parameter $lambda$, and since $E[X] = frac1lambda$,



beginalign
f(x) &= frac1mue^-frac1mux \
F(x) &= 1-e^-frac1mux
endalign



Year $Y$ can be odd if it's the $1$st year ($12$ months), $3$rd year ($36$ months) and so on.



I'm having troubles constructing the $P(Y= mathrmodd)$ model here, as I don't know how to put together the infinite odd years described above.






probability exponential-distribution




probability exponential-distribution






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 15 at 1:13









Rócherz

2,9863821




2,9863821










asked Mar 15 at 1:01









PTNPTN

1426




1426







  • 2




    $begingroup$
    If $X$ is the random month in which the lightbulb kaputs, and $Y$ is the year of that happening, then $Y$ is odd whenever $X$ is between $1$ and $12$, or $25$ and $36$, or $49$ and $60$, and so on.
    $endgroup$
    – Rócherz
    Mar 15 at 1:21












  • 2




    $begingroup$
    If $X$ is the random month in which the lightbulb kaputs, and $Y$ is the year of that happening, then $Y$ is odd whenever $X$ is between $1$ and $12$, or $25$ and $36$, or $49$ and $60$, and so on.
    $endgroup$
    – Rócherz
    Mar 15 at 1:21







2




2




$begingroup$
If $X$ is the random month in which the lightbulb kaputs, and $Y$ is the year of that happening, then $Y$ is odd whenever $X$ is between $1$ and $12$, or $25$ and $36$, or $49$ and $60$, and so on.
$endgroup$
– Rócherz
Mar 15 at 1:21




$begingroup$
If $X$ is the random month in which the lightbulb kaputs, and $Y$ is the year of that happening, then $Y$ is odd whenever $X$ is between $1$ and $12$, or $25$ and $36$, or $49$ and $60$, and so on.
$endgroup$
– Rócherz
Mar 15 at 1:21










1 Answer
1






active

oldest

votes


















1












$begingroup$

There's only one parameter in this exponential model – $lambda=frac1mu$. Thus
$$F_Y(y)=1-e^-lambda x$$
$$P(12k<Y<12(k+1))=F_Y(12(k+1))-F_Y(12k)=1-e^-12lambda(k+1)-1+e^-12lambda k$$
$$=-e^-12lambda(k+1)+e^-12lambda k=e^-12lambda k(1-e^-12lambda)$$
Then the probability the bulb fails in an odd year is an infinite sum with $k=2n=0,2,4dots$:
$$sum_n=0^infty e^-24lambda n(1-e^-12lambda)$$
$$=(1-e^-12lambda)sum_n=0^infty(e^-24lambda)^n=frac1-e^-12lambda1-e^-24lambda$$
$$=frac1-e^-12/mu1-e^-24/mu$$
(We can derive this faster by using the memorylessness of the exponential distribution; in each 24-month cycle the probability of the bulb failing in the first half of that cycle given that it survives to the start of that cycle is constant, and equal to $fracF_Y(12)F_Y(24)$.)






share|cite|improve this answer











$endgroup$












  • $begingroup$
    So $k$ here has to be even right?
    $endgroup$
    – PTN
    Mar 15 at 1:46










  • $begingroup$
    @PTN Yes. $$
    $endgroup$
    – Parcly Taxel
    Mar 15 at 1:46










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1 Answer
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active

oldest

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1 Answer
1






active

oldest

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active

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active

oldest

votes









1












$begingroup$

There's only one parameter in this exponential model – $lambda=frac1mu$. Thus
$$F_Y(y)=1-e^-lambda x$$
$$P(12k<Y<12(k+1))=F_Y(12(k+1))-F_Y(12k)=1-e^-12lambda(k+1)-1+e^-12lambda k$$
$$=-e^-12lambda(k+1)+e^-12lambda k=e^-12lambda k(1-e^-12lambda)$$
Then the probability the bulb fails in an odd year is an infinite sum with $k=2n=0,2,4dots$:
$$sum_n=0^infty e^-24lambda n(1-e^-12lambda)$$
$$=(1-e^-12lambda)sum_n=0^infty(e^-24lambda)^n=frac1-e^-12lambda1-e^-24lambda$$
$$=frac1-e^-12/mu1-e^-24/mu$$
(We can derive this faster by using the memorylessness of the exponential distribution; in each 24-month cycle the probability of the bulb failing in the first half of that cycle given that it survives to the start of that cycle is constant, and equal to $fracF_Y(12)F_Y(24)$.)






share|cite|improve this answer











$endgroup$












  • $begingroup$
    So $k$ here has to be even right?
    $endgroup$
    – PTN
    Mar 15 at 1:46










  • $begingroup$
    @PTN Yes. $$
    $endgroup$
    – Parcly Taxel
    Mar 15 at 1:46















1












$begingroup$

There's only one parameter in this exponential model – $lambda=frac1mu$. Thus
$$F_Y(y)=1-e^-lambda x$$
$$P(12k<Y<12(k+1))=F_Y(12(k+1))-F_Y(12k)=1-e^-12lambda(k+1)-1+e^-12lambda k$$
$$=-e^-12lambda(k+1)+e^-12lambda k=e^-12lambda k(1-e^-12lambda)$$
Then the probability the bulb fails in an odd year is an infinite sum with $k=2n=0,2,4dots$:
$$sum_n=0^infty e^-24lambda n(1-e^-12lambda)$$
$$=(1-e^-12lambda)sum_n=0^infty(e^-24lambda)^n=frac1-e^-12lambda1-e^-24lambda$$
$$=frac1-e^-12/mu1-e^-24/mu$$
(We can derive this faster by using the memorylessness of the exponential distribution; in each 24-month cycle the probability of the bulb failing in the first half of that cycle given that it survives to the start of that cycle is constant, and equal to $fracF_Y(12)F_Y(24)$.)






share|cite|improve this answer











$endgroup$












  • $begingroup$
    So $k$ here has to be even right?
    $endgroup$
    – PTN
    Mar 15 at 1:46










  • $begingroup$
    @PTN Yes. $$
    $endgroup$
    – Parcly Taxel
    Mar 15 at 1:46













1












1








1





$begingroup$

There's only one parameter in this exponential model – $lambda=frac1mu$. Thus
$$F_Y(y)=1-e^-lambda x$$
$$P(12k<Y<12(k+1))=F_Y(12(k+1))-F_Y(12k)=1-e^-12lambda(k+1)-1+e^-12lambda k$$
$$=-e^-12lambda(k+1)+e^-12lambda k=e^-12lambda k(1-e^-12lambda)$$
Then the probability the bulb fails in an odd year is an infinite sum with $k=2n=0,2,4dots$:
$$sum_n=0^infty e^-24lambda n(1-e^-12lambda)$$
$$=(1-e^-12lambda)sum_n=0^infty(e^-24lambda)^n=frac1-e^-12lambda1-e^-24lambda$$
$$=frac1-e^-12/mu1-e^-24/mu$$
(We can derive this faster by using the memorylessness of the exponential distribution; in each 24-month cycle the probability of the bulb failing in the first half of that cycle given that it survives to the start of that cycle is constant, and equal to $fracF_Y(12)F_Y(24)$.)






share|cite|improve this answer











$endgroup$



There's only one parameter in this exponential model – $lambda=frac1mu$. Thus
$$F_Y(y)=1-e^-lambda x$$
$$P(12k<Y<12(k+1))=F_Y(12(k+1))-F_Y(12k)=1-e^-12lambda(k+1)-1+e^-12lambda k$$
$$=-e^-12lambda(k+1)+e^-12lambda k=e^-12lambda k(1-e^-12lambda)$$
Then the probability the bulb fails in an odd year is an infinite sum with $k=2n=0,2,4dots$:
$$sum_n=0^infty e^-24lambda n(1-e^-12lambda)$$
$$=(1-e^-12lambda)sum_n=0^infty(e^-24lambda)^n=frac1-e^-12lambda1-e^-24lambda$$
$$=frac1-e^-12/mu1-e^-24/mu$$
(We can derive this faster by using the memorylessness of the exponential distribution; in each 24-month cycle the probability of the bulb failing in the first half of that cycle given that it survives to the start of that cycle is constant, and equal to $fracF_Y(12)F_Y(24)$.)







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Mar 15 at 1:29

























answered Mar 15 at 1:23









Parcly TaxelParcly Taxel

44.7k1376109




44.7k1376109











  • $begingroup$
    So $k$ here has to be even right?
    $endgroup$
    – PTN
    Mar 15 at 1:46










  • $begingroup$
    @PTN Yes. $$
    $endgroup$
    – Parcly Taxel
    Mar 15 at 1:46
















  • $begingroup$
    So $k$ here has to be even right?
    $endgroup$
    – PTN
    Mar 15 at 1:46










  • $begingroup$
    @PTN Yes. $$
    $endgroup$
    – Parcly Taxel
    Mar 15 at 1:46















$begingroup$
So $k$ here has to be even right?
$endgroup$
– PTN
Mar 15 at 1:46




$begingroup$
So $k$ here has to be even right?
$endgroup$
– PTN
Mar 15 at 1:46












$begingroup$
@PTN Yes. $$
$endgroup$
– Parcly Taxel
Mar 15 at 1:46




$begingroup$
@PTN Yes. $$
$endgroup$
– Parcly Taxel
Mar 15 at 1:46

















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