Understanding $P[X gt alpha] = P[e^theta X gt e^theta alpha]$
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In Chernof Bound, it is expressed $P[X gt alpha] = P[e^theta X gt e^theta alpha]$ for given $M_X(theta) = E[e^theta X]$.
I am not getting how we can equal $P[X gt alpha] = P[e^theta X gt e^theta alpha]$.
Can you clarify this?
probability random-variables
edited Aug 28 at 19:43
BruceET
33.7k71440
asked Aug 28 at 17:27
Malintha
1527
add a comment |Â
up vote
0
down vote
favorite
In Chernof Bound, it is expressed $P[X gt alpha] = P[e^theta X gt e^theta alpha]$ for given $M_X(theta) = E[e^theta X]$.
I am not getting how we can equal $P[X gt alpha] = P[e^theta X gt e^theta alpha]$.
Can you clarify this?
probability random-variables
edited Aug 28 at 19:43
BruceET
33.7k71440
asked Aug 28 at 17:27
Malintha
1527
2
Assuming $theta$ is positive, we have $P(X>alpha)=P(theta X>thetaalpha)$. Now use the fact that $e^x$ is an increasing function of $x$.
– StubbornAtom
Aug 28 at 17:35
2
If $f$ is increasing, then $[X > alpha] = [f(X) > f(alpha)]$.
– copper.hat
Aug 28 at 17:55
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
In Chernof Bound, it is expressed $P[X gt alpha] = P[e^theta X gt e^theta alpha]$ for given $M_X(theta) = E[e^theta X]$.
I am not getting how we can equal $P[X gt alpha] = P[e^theta X gt e^theta alpha]$.
Can you clarify this?
probability random-variables
edited Aug 28 at 19:43
BruceET
33.7k71440
asked Aug 28 at 17:27
Malintha
1527
In Chernof Bound, it is expressed $P[X gt alpha] = P[e^theta X gt e^theta alpha]$ for given $M_X(theta) = E[e^theta X]$.
I am not getting how we can equal $P[X gt alpha] = P[e^theta X gt e^theta alpha]$.
Can you clarify this?
probability random-variables
edited Aug 28 at 19:43
BruceET
33.7k71440
asked Aug 28 at 17:27
Malintha
1527
edited Aug 28 at 19:43
BruceET
33.7k71440
edited Aug 28 at 19:43
BruceET
33.7k71440
edited Aug 28 at 19:43
BruceET
33.7k71440
33.7k71440
asked Aug 28 at 17:27
Malintha
1527
asked Aug 28 at 17:27
Malintha
1527
asked Aug 28 at 17:27
Malintha
1527
1527
2
Assuming $theta$ is positive, we have $P(X>alpha)=P(theta X>thetaalpha)$. Now use the fact that $e^x$ is an increasing function of $x$.
– StubbornAtom
Aug 28 at 17:35
2
If $f$ is increasing, then $[X > alpha] = [f(X) > f(alpha)]$.
– copper.hat
Aug 28 at 17:55
add a comment |Â
2
Assuming $theta$ is positive, we have $P(X>alpha)=P(theta X>thetaalpha)$. Now use the fact that $e^x$ is an increasing function of $x$.
– StubbornAtom
Aug 28 at 17:35
2
If $f$ is increasing, then $[X > alpha] = [f(X) > f(alpha)]$.
– copper.hat
Aug 28 at 17:55
2
2
Assuming $theta$ is positive, we have $P(X>alpha)=P(theta X>thetaalpha)$. Now use the fact that $e^x$ is an increasing function of $x$.
– StubbornAtom
Aug 28 at 17:35
Assuming $theta$ is positive, we have $P(X>alpha)=P(theta X>thetaalpha)$. Now use the fact that $e^x$ is an increasing function of $x$.
– StubbornAtom
Aug 28 at 17:35
2
2
If $f$ is increasing, then $[X > alpha] = [f(X) > f(alpha)]$.
– copper.hat
Aug 28 at 17:55
If $f$ is increasing, then $[X > alpha] = [f(X) > f(alpha)]$.
– copper.hat
Aug 28 at 17:55
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
The fundamental observations are that
beginalign*
& X > a \
iff & theta X > theta a qquad text(assuming $theta > 0$) \
iff & e^theta X > e^theta a
endalign*
because $e^x$ is an increasing function. Hence, the events $X > a$ and $e^theta X > e^theta a$ are the same event.
It may be instructive to compare a statement that would be proved similarly: $mathbb P(X > 3) = mathbb P(5X + 2 > 17)$. You're permitted to work algebraically on the underlying inequalities, which is what's happening here.
answered Aug 28 at 22:01
Aaron Montgomery
4,342423
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
The fundamental observations are that
beginalign*
& X > a \
iff & theta X > theta a qquad text(assuming $theta > 0$) \
iff & e^theta X > e^theta a
endalign*
because $e^x$ is an increasing function. Hence, the events $X > a$ and $e^theta X > e^theta a$ are the same event.
It may be instructive to compare a statement that would be proved similarly: $mathbb P(X > 3) = mathbb P(5X + 2 > 17)$. You're permitted to work algebraically on the underlying inequalities, which is what's happening here.
answered Aug 28 at 22:01
Aaron Montgomery
4,342423
add a comment |Â
up vote
2
down vote
accepted
The fundamental observations are that
beginalign*
& X > a \
iff & theta X > theta a qquad text(assuming $theta > 0$) \
iff & e^theta X > e^theta a
endalign*
because $e^x$ is an increasing function. Hence, the events $X > a$ and $e^theta X > e^theta a$ are the same event.
It may be instructive to compare a statement that would be proved similarly: $mathbb P(X > 3) = mathbb P(5X + 2 > 17)$. You're permitted to work algebraically on the underlying inequalities, which is what's happening here.
answered Aug 28 at 22:01
Aaron Montgomery
4,342423
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
The fundamental observations are that
beginalign*
& X > a \
iff & theta X > theta a qquad text(assuming $theta > 0$) \
iff & e^theta X > e^theta a
endalign*
because $e^x$ is an increasing function. Hence, the events $X > a$ and $e^theta X > e^theta a$ are the same event.
It may be instructive to compare a statement that would be proved similarly: $mathbb P(X > 3) = mathbb P(5X + 2 > 17)$. You're permitted to work algebraically on the underlying inequalities, which is what's happening here.
answered Aug 28 at 22:01
Aaron Montgomery
4,342423
The fundamental observations are that
beginalign*
& X > a \
iff & theta X > theta a qquad text(assuming $theta > 0$) \
iff & e^theta X > e^theta a
endalign*
because $e^x$ is an increasing function. Hence, the events $X > a$ and $e^theta X > e^theta a$ are the same event.
It may be instructive to compare a statement that would be proved similarly: $mathbb P(X > 3) = mathbb P(5X + 2 > 17)$. You're permitted to work algebraically on the underlying inequalities, which is what's happening here.
answered Aug 28 at 22:01
Aaron Montgomery
4,342423
answered Aug 28 at 22:01
Aaron Montgomery
4,342423
answered Aug 28 at 22:01
Aaron Montgomery
4,342423
answered Aug 28 at 22:01
Aaron Montgomery
4,342423
4,342423
add a comment |Â
add a comment |Â
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2
Assuming $theta$ is positive, we have $P(X>alpha)=P(theta X>thetaalpha)$. Now use the fact that $e^x$ is an increasing function of $x$.
– StubbornAtom
Aug 28 at 17:35
2
If $f$ is increasing, then $[X > alpha] = [f(X) > f(alpha)]$.
– copper.hat
Aug 28 at 17:55