Is $T(X)$ a sufficient statistic for $lambda$?












0












$begingroup$


Let $X$ be a sample (size n = 1) from the exponential distribution, which has the pdf $$f(x;lambda) = lambda exp(-lambda x)$$ where $lambda$ is an unknown parameter. Let's define a statistic as



$$T(X)= left{ begin{array}{lcc}
1 & X > 2 \
\ 0 &mbox{ otherwise} \
\
end{array}
right.$$



Is $T(X)$ a sufficient statistic for $lambda$?



I honestly have no idea on how to approach this problem. Any suggestions?










share|cite|improve this question









$endgroup$












  • $begingroup$
    The classic way to prove something about sufficiency is to try to use the factorization theorem. If I give you the hint that $T(X) = 1_{X > 2}$, does that help?
    $endgroup$
    – OldGodzilla
    Jan 16 at 20:47










  • $begingroup$
    @OldGodzilla, I know $T(X)=1$ but how do I proceed from there?
    $endgroup$
    – Lady
    Jan 16 at 21:37










  • $begingroup$
    It's not that $T(X) = 1$, it's that $T(X)$ is the indicator of an event. If $T$ is sufficient for $lambda$, then distribution can be factored into the product of two parts: one depending just on $x$ and one depending only on $lambda$ and $T(X)$.
    $endgroup$
    – OldGodzilla
    Jan 16 at 21:45
















0












$begingroup$


Let $X$ be a sample (size n = 1) from the exponential distribution, which has the pdf $$f(x;lambda) = lambda exp(-lambda x)$$ where $lambda$ is an unknown parameter. Let's define a statistic as



$$T(X)= left{ begin{array}{lcc}
1 & X > 2 \
\ 0 &mbox{ otherwise} \
\
end{array}
right.$$



Is $T(X)$ a sufficient statistic for $lambda$?



I honestly have no idea on how to approach this problem. Any suggestions?










share|cite|improve this question









$endgroup$












  • $begingroup$
    The classic way to prove something about sufficiency is to try to use the factorization theorem. If I give you the hint that $T(X) = 1_{X > 2}$, does that help?
    $endgroup$
    – OldGodzilla
    Jan 16 at 20:47










  • $begingroup$
    @OldGodzilla, I know $T(X)=1$ but how do I proceed from there?
    $endgroup$
    – Lady
    Jan 16 at 21:37










  • $begingroup$
    It's not that $T(X) = 1$, it's that $T(X)$ is the indicator of an event. If $T$ is sufficient for $lambda$, then distribution can be factored into the product of two parts: one depending just on $x$ and one depending only on $lambda$ and $T(X)$.
    $endgroup$
    – OldGodzilla
    Jan 16 at 21:45














0












0








0


0



$begingroup$


Let $X$ be a sample (size n = 1) from the exponential distribution, which has the pdf $$f(x;lambda) = lambda exp(-lambda x)$$ where $lambda$ is an unknown parameter. Let's define a statistic as



$$T(X)= left{ begin{array}{lcc}
1 & X > 2 \
\ 0 &mbox{ otherwise} \
\
end{array}
right.$$



Is $T(X)$ a sufficient statistic for $lambda$?



I honestly have no idea on how to approach this problem. Any suggestions?










share|cite|improve this question









$endgroup$




Let $X$ be a sample (size n = 1) from the exponential distribution, which has the pdf $$f(x;lambda) = lambda exp(-lambda x)$$ where $lambda$ is an unknown parameter. Let's define a statistic as



$$T(X)= left{ begin{array}{lcc}
1 & X > 2 \
\ 0 &mbox{ otherwise} \
\
end{array}
right.$$



Is $T(X)$ a sufficient statistic for $lambda$?



I honestly have no idea on how to approach this problem. Any suggestions?







statistics statistical-inference






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 16 at 19:50









LadyLady

1298




1298












  • $begingroup$
    The classic way to prove something about sufficiency is to try to use the factorization theorem. If I give you the hint that $T(X) = 1_{X > 2}$, does that help?
    $endgroup$
    – OldGodzilla
    Jan 16 at 20:47










  • $begingroup$
    @OldGodzilla, I know $T(X)=1$ but how do I proceed from there?
    $endgroup$
    – Lady
    Jan 16 at 21:37










  • $begingroup$
    It's not that $T(X) = 1$, it's that $T(X)$ is the indicator of an event. If $T$ is sufficient for $lambda$, then distribution can be factored into the product of two parts: one depending just on $x$ and one depending only on $lambda$ and $T(X)$.
    $endgroup$
    – OldGodzilla
    Jan 16 at 21:45


















  • $begingroup$
    The classic way to prove something about sufficiency is to try to use the factorization theorem. If I give you the hint that $T(X) = 1_{X > 2}$, does that help?
    $endgroup$
    – OldGodzilla
    Jan 16 at 20:47










  • $begingroup$
    @OldGodzilla, I know $T(X)=1$ but how do I proceed from there?
    $endgroup$
    – Lady
    Jan 16 at 21:37










  • $begingroup$
    It's not that $T(X) = 1$, it's that $T(X)$ is the indicator of an event. If $T$ is sufficient for $lambda$, then distribution can be factored into the product of two parts: one depending just on $x$ and one depending only on $lambda$ and $T(X)$.
    $endgroup$
    – OldGodzilla
    Jan 16 at 21:45
















$begingroup$
The classic way to prove something about sufficiency is to try to use the factorization theorem. If I give you the hint that $T(X) = 1_{X > 2}$, does that help?
$endgroup$
– OldGodzilla
Jan 16 at 20:47




$begingroup$
The classic way to prove something about sufficiency is to try to use the factorization theorem. If I give you the hint that $T(X) = 1_{X > 2}$, does that help?
$endgroup$
– OldGodzilla
Jan 16 at 20:47












$begingroup$
@OldGodzilla, I know $T(X)=1$ but how do I proceed from there?
$endgroup$
– Lady
Jan 16 at 21:37




$begingroup$
@OldGodzilla, I know $T(X)=1$ but how do I proceed from there?
$endgroup$
– Lady
Jan 16 at 21:37












$begingroup$
It's not that $T(X) = 1$, it's that $T(X)$ is the indicator of an event. If $T$ is sufficient for $lambda$, then distribution can be factored into the product of two parts: one depending just on $x$ and one depending only on $lambda$ and $T(X)$.
$endgroup$
– OldGodzilla
Jan 16 at 21:45




$begingroup$
It's not that $T(X) = 1$, it's that $T(X)$ is the indicator of an event. If $T$ is sufficient for $lambda$, then distribution can be factored into the product of two parts: one depending just on $x$ and one depending only on $lambda$ and $T(X)$.
$endgroup$
– OldGodzilla
Jan 16 at 21:45










1 Answer
1






active

oldest

votes


















0












$begingroup$

Use the definition of sufficiency to prove that a statistic is not sufficient; the Factorisation theorem comes in handy when you are proving that a statistic is sufficient.



For any $xgeqslant 0$, the distribution function of $Xmid T=1$ is



begin{align}
P(Xleqslant xmid T=1)&=begin{cases}frac{P(Xleqslant x)}{P(T=1)}&,text{ if }T=1 \ quad 0 &,text{ otherwise }end{cases}
\&=begin{cases}frac{P(Xleqslant x)}{P(X>2)}&,text{ if }T=1 \ quad 0&,text{ otherwise }end{cases}
end{align}



Can you see now whether the conditional distribution $Xmid T$ depends on $lambda$ or not? From this information, conclude about the sufficiency of $T$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    for both cases the conditional distribution depends on $lambda$. Would that be sufficient to say that the statistic is not sufficient then?
    $endgroup$
    – Lady
    Jan 17 at 12:06










  • $begingroup$
    @Lady Yes, that's what the definition of sufficiency says.
    $endgroup$
    – StubbornAtom
    Jan 17 at 12:08










  • $begingroup$
    but why is x defined to be greater than or equal to zero in your definition?
    $endgroup$
    – Lady
    Jan 17 at 14:16










  • $begingroup$
    @Lady Because that is the support of your exponential distribution.
    $endgroup$
    – StubbornAtom
    Jan 17 at 14:19










  • $begingroup$
    why use $P(Xleqslant xmid T=1)$ ?
    $endgroup$
    – Lady
    Jan 20 at 2:33












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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









0












$begingroup$

Use the definition of sufficiency to prove that a statistic is not sufficient; the Factorisation theorem comes in handy when you are proving that a statistic is sufficient.



For any $xgeqslant 0$, the distribution function of $Xmid T=1$ is



begin{align}
P(Xleqslant xmid T=1)&=begin{cases}frac{P(Xleqslant x)}{P(T=1)}&,text{ if }T=1 \ quad 0 &,text{ otherwise }end{cases}
\&=begin{cases}frac{P(Xleqslant x)}{P(X>2)}&,text{ if }T=1 \ quad 0&,text{ otherwise }end{cases}
end{align}



Can you see now whether the conditional distribution $Xmid T$ depends on $lambda$ or not? From this information, conclude about the sufficiency of $T$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    for both cases the conditional distribution depends on $lambda$. Would that be sufficient to say that the statistic is not sufficient then?
    $endgroup$
    – Lady
    Jan 17 at 12:06










  • $begingroup$
    @Lady Yes, that's what the definition of sufficiency says.
    $endgroup$
    – StubbornAtom
    Jan 17 at 12:08










  • $begingroup$
    but why is x defined to be greater than or equal to zero in your definition?
    $endgroup$
    – Lady
    Jan 17 at 14:16










  • $begingroup$
    @Lady Because that is the support of your exponential distribution.
    $endgroup$
    – StubbornAtom
    Jan 17 at 14:19










  • $begingroup$
    why use $P(Xleqslant xmid T=1)$ ?
    $endgroup$
    – Lady
    Jan 20 at 2:33
















0












$begingroup$

Use the definition of sufficiency to prove that a statistic is not sufficient; the Factorisation theorem comes in handy when you are proving that a statistic is sufficient.



For any $xgeqslant 0$, the distribution function of $Xmid T=1$ is



begin{align}
P(Xleqslant xmid T=1)&=begin{cases}frac{P(Xleqslant x)}{P(T=1)}&,text{ if }T=1 \ quad 0 &,text{ otherwise }end{cases}
\&=begin{cases}frac{P(Xleqslant x)}{P(X>2)}&,text{ if }T=1 \ quad 0&,text{ otherwise }end{cases}
end{align}



Can you see now whether the conditional distribution $Xmid T$ depends on $lambda$ or not? From this information, conclude about the sufficiency of $T$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    for both cases the conditional distribution depends on $lambda$. Would that be sufficient to say that the statistic is not sufficient then?
    $endgroup$
    – Lady
    Jan 17 at 12:06










  • $begingroup$
    @Lady Yes, that's what the definition of sufficiency says.
    $endgroup$
    – StubbornAtom
    Jan 17 at 12:08










  • $begingroup$
    but why is x defined to be greater than or equal to zero in your definition?
    $endgroup$
    – Lady
    Jan 17 at 14:16










  • $begingroup$
    @Lady Because that is the support of your exponential distribution.
    $endgroup$
    – StubbornAtom
    Jan 17 at 14:19










  • $begingroup$
    why use $P(Xleqslant xmid T=1)$ ?
    $endgroup$
    – Lady
    Jan 20 at 2:33














0












0








0





$begingroup$

Use the definition of sufficiency to prove that a statistic is not sufficient; the Factorisation theorem comes in handy when you are proving that a statistic is sufficient.



For any $xgeqslant 0$, the distribution function of $Xmid T=1$ is



begin{align}
P(Xleqslant xmid T=1)&=begin{cases}frac{P(Xleqslant x)}{P(T=1)}&,text{ if }T=1 \ quad 0 &,text{ otherwise }end{cases}
\&=begin{cases}frac{P(Xleqslant x)}{P(X>2)}&,text{ if }T=1 \ quad 0&,text{ otherwise }end{cases}
end{align}



Can you see now whether the conditional distribution $Xmid T$ depends on $lambda$ or not? From this information, conclude about the sufficiency of $T$.






share|cite|improve this answer









$endgroup$



Use the definition of sufficiency to prove that a statistic is not sufficient; the Factorisation theorem comes in handy when you are proving that a statistic is sufficient.



For any $xgeqslant 0$, the distribution function of $Xmid T=1$ is



begin{align}
P(Xleqslant xmid T=1)&=begin{cases}frac{P(Xleqslant x)}{P(T=1)}&,text{ if }T=1 \ quad 0 &,text{ otherwise }end{cases}
\&=begin{cases}frac{P(Xleqslant x)}{P(X>2)}&,text{ if }T=1 \ quad 0&,text{ otherwise }end{cases}
end{align}



Can you see now whether the conditional distribution $Xmid T$ depends on $lambda$ or not? From this information, conclude about the sufficiency of $T$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 17 at 12:00









StubbornAtomStubbornAtom

6,37831440




6,37831440












  • $begingroup$
    for both cases the conditional distribution depends on $lambda$. Would that be sufficient to say that the statistic is not sufficient then?
    $endgroup$
    – Lady
    Jan 17 at 12:06










  • $begingroup$
    @Lady Yes, that's what the definition of sufficiency says.
    $endgroup$
    – StubbornAtom
    Jan 17 at 12:08










  • $begingroup$
    but why is x defined to be greater than or equal to zero in your definition?
    $endgroup$
    – Lady
    Jan 17 at 14:16










  • $begingroup$
    @Lady Because that is the support of your exponential distribution.
    $endgroup$
    – StubbornAtom
    Jan 17 at 14:19










  • $begingroup$
    why use $P(Xleqslant xmid T=1)$ ?
    $endgroup$
    – Lady
    Jan 20 at 2:33


















  • $begingroup$
    for both cases the conditional distribution depends on $lambda$. Would that be sufficient to say that the statistic is not sufficient then?
    $endgroup$
    – Lady
    Jan 17 at 12:06










  • $begingroup$
    @Lady Yes, that's what the definition of sufficiency says.
    $endgroup$
    – StubbornAtom
    Jan 17 at 12:08










  • $begingroup$
    but why is x defined to be greater than or equal to zero in your definition?
    $endgroup$
    – Lady
    Jan 17 at 14:16










  • $begingroup$
    @Lady Because that is the support of your exponential distribution.
    $endgroup$
    – StubbornAtom
    Jan 17 at 14:19










  • $begingroup$
    why use $P(Xleqslant xmid T=1)$ ?
    $endgroup$
    – Lady
    Jan 20 at 2:33
















$begingroup$
for both cases the conditional distribution depends on $lambda$. Would that be sufficient to say that the statistic is not sufficient then?
$endgroup$
– Lady
Jan 17 at 12:06




$begingroup$
for both cases the conditional distribution depends on $lambda$. Would that be sufficient to say that the statistic is not sufficient then?
$endgroup$
– Lady
Jan 17 at 12:06












$begingroup$
@Lady Yes, that's what the definition of sufficiency says.
$endgroup$
– StubbornAtom
Jan 17 at 12:08




$begingroup$
@Lady Yes, that's what the definition of sufficiency says.
$endgroup$
– StubbornAtom
Jan 17 at 12:08












$begingroup$
but why is x defined to be greater than or equal to zero in your definition?
$endgroup$
– Lady
Jan 17 at 14:16




$begingroup$
but why is x defined to be greater than or equal to zero in your definition?
$endgroup$
– Lady
Jan 17 at 14:16












$begingroup$
@Lady Because that is the support of your exponential distribution.
$endgroup$
– StubbornAtom
Jan 17 at 14:19




$begingroup$
@Lady Because that is the support of your exponential distribution.
$endgroup$
– StubbornAtom
Jan 17 at 14:19












$begingroup$
why use $P(Xleqslant xmid T=1)$ ?
$endgroup$
– Lady
Jan 20 at 2:33




$begingroup$
why use $P(Xleqslant xmid T=1)$ ?
$endgroup$
– Lady
Jan 20 at 2:33


















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