Is $T(X)$ a sufficient statistic for $lambda$?
$begingroup$
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?
statistics statistical-inference
asked Jan 16 at 19:50
LadyLady
1298
$endgroup$
add a comment |
$begingroup$
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?
statistics statistical-inference
asked Jan 16 at 19:50
LadyLady
1298
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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?
statistics statistical-inference
asked Jan 16 at 19:50
LadyLady
1298
$endgroup$
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
statistics statistical-inference
asked Jan 16 at 19:50
LadyLady
1298
asked Jan 16 at 19:50
LadyLady
1298
asked Jan 16 at 19:50
LadyLady
1298
asked Jan 16 at 19:50
LadyLady
1298
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
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$begingroup$
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 $T$.
answered Jan 17 at 12:00
StubbornAtomStubbornAtom
6,37831440
$endgroup$
$begingroup$
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
|
show 4 more comments
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$begingroup$
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 $T$.
answered Jan 17 at 12:00
StubbornAtomStubbornAtom
6,37831440
$endgroup$
$begingroup$
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
|
show 4 more comments
$begingroup$
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 $T$.
answered Jan 17 at 12:00
StubbornAtomStubbornAtom
6,37831440
$endgroup$
$begingroup$
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
|
show 4 more comments
$begingroup$
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 $T$.
answered Jan 17 at 12:00
StubbornAtomStubbornAtom
6,37831440
$endgroup$
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 $T$.
answered Jan 17 at 12:00
StubbornAtomStubbornAtom
6,37831440
answered Jan 17 at 12:00
StubbornAtomStubbornAtom
6,37831440
answered Jan 17 at 12:00
StubbornAtomStubbornAtom
6,37831440
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
|
show 4 more comments
$begingroup$
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
|
show 4 more comments
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$begingroup$
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