Chebyshev variant
Show that $P(a<X<b)geq 1-frac{sigma^2+(mu-frac{a+b}{2})^2}{(frac{b-a}{2})^2}$ where $X$ is a random variable with mean $mu$ and variance $sigma^2$.
I'm having a hard time with this question. I was trying to think of a new variable $X=Y+k$ to so that on $P(a-k<Y<b-k)$ you can just directly apply chebyshev, but I now realize that just shifts everything over so it doesn't help. I was thinking of splitting the interval into $a,mu+mu-a$ and $mu+mu-a,b$, but then I have no idea what to do with the second interval.
It kind of looks to me like the right side somehow combines the inequality for two different variables, but I have no idea how I would get a variable with variance $(mu-frac{a+b}{2})^2$.
Any ideas for how this problem is supposed to be solved?
probability
asked Nov 30 at 5:05
Miles Johnson
1928
add a comment |
Show that $P(a<X<b)geq 1-frac{sigma^2+(mu-frac{a+b}{2})^2}{(frac{b-a}{2})^2}$ where $X$ is a random variable with mean $mu$ and variance $sigma^2$.
I'm having a hard time with this question. I was trying to think of a new variable $X=Y+k$ to so that on $P(a-k<Y<b-k)$ you can just directly apply chebyshev, but I now realize that just shifts everything over so it doesn't help. I was thinking of splitting the interval into $a,mu+mu-a$ and $mu+mu-a,b$, but then I have no idea what to do with the second interval.
It kind of looks to me like the right side somehow combines the inequality for two different variables, but I have no idea how I would get a variable with variance $(mu-frac{a+b}{2})^2$.
Any ideas for how this problem is supposed to be solved?
probability
asked Nov 30 at 5:05
Miles Johnson
1928
add a comment |
Show that $P(a<X<b)geq 1-frac{sigma^2+(mu-frac{a+b}{2})^2}{(frac{b-a}{2})^2}$ where $X$ is a random variable with mean $mu$ and variance $sigma^2$.
I'm having a hard time with this question. I was trying to think of a new variable $X=Y+k$ to so that on $P(a-k<Y<b-k)$ you can just directly apply chebyshev, but I now realize that just shifts everything over so it doesn't help. I was thinking of splitting the interval into $a,mu+mu-a$ and $mu+mu-a,b$, but then I have no idea what to do with the second interval.
It kind of looks to me like the right side somehow combines the inequality for two different variables, but I have no idea how I would get a variable with variance $(mu-frac{a+b}{2})^2$.
Any ideas for how this problem is supposed to be solved?
probability
asked Nov 30 at 5:05
Miles Johnson
1928
Show that $P(a<X<b)geq 1-frac{sigma^2+(mu-frac{a+b}{2})^2}{(frac{b-a}{2})^2}$ where $X$ is a random variable with mean $mu$ and variance $sigma^2$.
I'm having a hard time with this question. I was trying to think of a new variable $X=Y+k$ to so that on $P(a-k<Y<b-k)$ you can just directly apply chebyshev, but I now realize that just shifts everything over so it doesn't help. I was thinking of splitting the interval into $a,mu+mu-a$ and $mu+mu-a,b$, but then I have no idea what to do with the second interval.
It kind of looks to me like the right side somehow combines the inequality for two different variables, but I have no idea how I would get a variable with variance $(mu-frac{a+b}{2})^2$.
Any ideas for how this problem is supposed to be solved?
probability
probability
asked Nov 30 at 5:05
Miles Johnson
1928
asked Nov 30 at 5:05
Miles Johnson
1928
asked Nov 30 at 5:05
Miles Johnson
1928
asked Nov 30 at 5:05
Miles Johnson
1928
asked Nov 30 at 5:05
Miles Johnson
1928
1928
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
You're on the right track with the approach $X=Y+k$. Try setting $k$ to be the midpoint of the interval $(a,b)$, so $k:=frac12(a+b)$, and then the event ${a<X<b}$ is the same as the event ${|Y|<frac12(b-a)}$. For brevity write $c:=frac12(b-a)$.The complement of this last event is then ${|Y|ge c}$, and its probability is
$$
Pleft(|Y|ge cright)=P(Y^2ge c^2)le frac{E(Y^2)}{c^2}
$$
by Markov's inequality. Finish off by computing $E(Y^2)=operatorname{Var}(Y)+[E(Y)]^2$.
answered Nov 30 at 5:23
grand_chat
20k11225
add a comment |
For $a<b$,
$$
mathsf{P}(a<X<b)=mathsf{P}(|X-(a+b)/2|<(b-a)/2).
$$
answered Nov 30 at 5:20
d.k.o.
8,492527
add a comment |
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2 Answers
2
active
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votes
2 Answers
2
active
oldest
votes
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oldest
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oldest
votes
You're on the right track with the approach $X=Y+k$. Try setting $k$ to be the midpoint of the interval $(a,b)$, so $k:=frac12(a+b)$, and then the event ${a<X<b}$ is the same as the event ${|Y|<frac12(b-a)}$. For brevity write $c:=frac12(b-a)$.The complement of this last event is then ${|Y|ge c}$, and its probability is
$$
Pleft(|Y|ge cright)=P(Y^2ge c^2)le frac{E(Y^2)}{c^2}
$$
by Markov's inequality. Finish off by computing $E(Y^2)=operatorname{Var}(Y)+[E(Y)]^2$.
answered Nov 30 at 5:23
grand_chat
20k11225
add a comment |
You're on the right track with the approach $X=Y+k$. Try setting $k$ to be the midpoint of the interval $(a,b)$, so $k:=frac12(a+b)$, and then the event ${a<X<b}$ is the same as the event ${|Y|<frac12(b-a)}$. For brevity write $c:=frac12(b-a)$.The complement of this last event is then ${|Y|ge c}$, and its probability is
$$
Pleft(|Y|ge cright)=P(Y^2ge c^2)le frac{E(Y^2)}{c^2}
$$
by Markov's inequality. Finish off by computing $E(Y^2)=operatorname{Var}(Y)+[E(Y)]^2$.
answered Nov 30 at 5:23
grand_chat
20k11225
add a comment |
You're on the right track with the approach $X=Y+k$. Try setting $k$ to be the midpoint of the interval $(a,b)$, so $k:=frac12(a+b)$, and then the event ${a<X<b}$ is the same as the event ${|Y|<frac12(b-a)}$. For brevity write $c:=frac12(b-a)$.The complement of this last event is then ${|Y|ge c}$, and its probability is
$$
Pleft(|Y|ge cright)=P(Y^2ge c^2)le frac{E(Y^2)}{c^2}
$$
by Markov's inequality. Finish off by computing $E(Y^2)=operatorname{Var}(Y)+[E(Y)]^2$.
answered Nov 30 at 5:23
grand_chat
20k11225
You're on the right track with the approach $X=Y+k$. Try setting $k$ to be the midpoint of the interval $(a,b)$, so $k:=frac12(a+b)$, and then the event ${a<X<b}$ is the same as the event ${|Y|<frac12(b-a)}$. For brevity write $c:=frac12(b-a)$.The complement of this last event is then ${|Y|ge c}$, and its probability is
$$
Pleft(|Y|ge cright)=P(Y^2ge c^2)le frac{E(Y^2)}{c^2}
$$
by Markov's inequality. Finish off by computing $E(Y^2)=operatorname{Var}(Y)+[E(Y)]^2$.
answered Nov 30 at 5:23
grand_chat
20k11225
edited Nov 30 at 5:31
answered Nov 30 at 5:23
grand_chat
20k11225
answered Nov 30 at 5:23
grand_chat
20k11225
answered Nov 30 at 5:23
grand_chat
20k11225
20k11225
add a comment |
add a comment |
For $a<b$,
$$
mathsf{P}(a<X<b)=mathsf{P}(|X-(a+b)/2|<(b-a)/2).
$$
answered Nov 30 at 5:20
d.k.o.
8,492527
add a comment |
For $a<b$,
$$
mathsf{P}(a<X<b)=mathsf{P}(|X-(a+b)/2|<(b-a)/2).
$$
answered Nov 30 at 5:20
d.k.o.
8,492527
add a comment |
For $a<b$,
$$
mathsf{P}(a<X<b)=mathsf{P}(|X-(a+b)/2|<(b-a)/2).
$$
answered Nov 30 at 5:20
d.k.o.
8,492527
For $a<b$,
$$
mathsf{P}(a<X<b)=mathsf{P}(|X-(a+b)/2|<(b-a)/2).
$$
answered Nov 30 at 5:20
d.k.o.
8,492527
answered Nov 30 at 5:20
d.k.o.
8,492527
answered Nov 30 at 5:20
d.k.o.
8,492527
answered Nov 30 at 5:20
d.k.o.
8,492527
8,492527
add a comment |
add a comment |
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