A random variable X has range from $(0,a)$. Show that $Var(X)le a^2/4$
A random variable X has range from $(0,a)$. Show that $Var(X)le a^2/4$.
The given hint is to show $E[X^2]le aE[X]$ first, and use that to show $Var(X)le a^2[beta(1-beta)]$, where $beta=E[X]/a$.
I don't know how to start. I've tried using Cheyshev's inequality but it doesn't work. Can anyone give me a hint how to start? Any concept I've been missing?
probability
asked Nov 27 '18 at 1:09
Yibei He
708
add a comment |
A random variable X has range from $(0,a)$. Show that $Var(X)le a^2/4$.
The given hint is to show $E[X^2]le aE[X]$ first, and use that to show $Var(X)le a^2[beta(1-beta)]$, where $beta=E[X]/a$.
I don't know how to start. I've tried using Cheyshev's inequality but it doesn't work. Can anyone give me a hint how to start? Any concept I've been missing?
probability
asked Nov 27 '18 at 1:09
Yibei He
708
is that all the information in the question?
– Siong Thye Goh
Nov 27 '18 at 1:12
@SiongThyeGoh I've edited some typos. And yes, that is all
– Yibei He
Nov 27 '18 at 1:19
add a comment |
A random variable X has range from $(0,a)$. Show that $Var(X)le a^2/4$.
The given hint is to show $E[X^2]le aE[X]$ first, and use that to show $Var(X)le a^2[beta(1-beta)]$, where $beta=E[X]/a$.
I don't know how to start. I've tried using Cheyshev's inequality but it doesn't work. Can anyone give me a hint how to start? Any concept I've been missing?
probability
asked Nov 27 '18 at 1:09
Yibei He
708
A random variable X has range from $(0,a)$. Show that $Var(X)le a^2/4$.
The given hint is to show $E[X^2]le aE[X]$ first, and use that to show $Var(X)le a^2[beta(1-beta)]$, where $beta=E[X]/a$.
I don't know how to start. I've tried using Cheyshev's inequality but it doesn't work. Can anyone give me a hint how to start? Any concept I've been missing?
probability
probability
asked Nov 27 '18 at 1:09
Yibei He
708
asked Nov 27 '18 at 1:09
Yibei He
708
edited Nov 27 '18 at 1:16
asked Nov 27 '18 at 1:09
Yibei He
708
asked Nov 27 '18 at 1:09
Yibei He
708
asked Nov 27 '18 at 1:09
Yibei He
708
708
is that all the information in the question?
– Siong Thye Goh
Nov 27 '18 at 1:12
@SiongThyeGoh I've edited some typos. And yes, that is all
– Yibei He
Nov 27 '18 at 1:19
add a comment |
is that all the information in the question?
– Siong Thye Goh
Nov 27 '18 at 1:12
@SiongThyeGoh I've edited some typos. And yes, that is all
– Yibei He
Nov 27 '18 at 1:19
is that all the information in the question?
– Siong Thye Goh
Nov 27 '18 at 1:12
is that all the information in the question?
– Siong Thye Goh
Nov 27 '18 at 1:12
@SiongThyeGoh I've edited some typos. And yes, that is all
– Yibei He
Nov 27 '18 at 1:19
@SiongThyeGoh I've edited some typos. And yes, that is all
– Yibei He
Nov 27 '18 at 1:19
add a comment |
2 Answers
2
active
oldest
votes
$$E[aX-X^2]=E[X(a-X)]ge 0$$
$$Var(X)=E[X^2]-E[X]^2le aE[X]-E[X]^2$$
answered Nov 27 '18 at 1:25
Siong Thye Goh
99.4k1464117
add a comment |
You know that $Var(X)=int_0^a(x-mu)^2dx=int_0^ax^2p(x)dx-left(int_0^axp(x)dxright)^2$, where
$p(x)geq0$ is the probability distribution and $mu=E(X)$. Hence,
$$
Var(X)leq aint_0^axp(x)dx-mu^2=mu(a-mu)=a^2beta(1-beta)=-a^2(beta^2-beta+1/4-1/4)=-a^2(beta-1/2)^2+a^2/4leq a^2/4
$$
since $0leq xleq a$.
answered Nov 27 '18 at 1:34
minmax
48518
add a comment |
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2 Answers
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2 Answers
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active
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$$E[aX-X^2]=E[X(a-X)]ge 0$$
$$Var(X)=E[X^2]-E[X]^2le aE[X]-E[X]^2$$
answered Nov 27 '18 at 1:25
Siong Thye Goh
99.4k1464117
add a comment |
$$E[aX-X^2]=E[X(a-X)]ge 0$$
$$Var(X)=E[X^2]-E[X]^2le aE[X]-E[X]^2$$
answered Nov 27 '18 at 1:25
Siong Thye Goh
99.4k1464117
add a comment |
$$E[aX-X^2]=E[X(a-X)]ge 0$$
$$Var(X)=E[X^2]-E[X]^2le aE[X]-E[X]^2$$
answered Nov 27 '18 at 1:25
Siong Thye Goh
99.4k1464117
$$E[aX-X^2]=E[X(a-X)]ge 0$$
$$Var(X)=E[X^2]-E[X]^2le aE[X]-E[X]^2$$
answered Nov 27 '18 at 1:25
Siong Thye Goh
99.4k1464117
answered Nov 27 '18 at 1:25
Siong Thye Goh
99.4k1464117
answered Nov 27 '18 at 1:25
Siong Thye Goh
99.4k1464117
answered Nov 27 '18 at 1:25
Siong Thye Goh
99.4k1464117
99.4k1464117
add a comment |
add a comment |
You know that $Var(X)=int_0^a(x-mu)^2dx=int_0^ax^2p(x)dx-left(int_0^axp(x)dxright)^2$, where
$p(x)geq0$ is the probability distribution and $mu=E(X)$. Hence,
$$
Var(X)leq aint_0^axp(x)dx-mu^2=mu(a-mu)=a^2beta(1-beta)=-a^2(beta^2-beta+1/4-1/4)=-a^2(beta-1/2)^2+a^2/4leq a^2/4
$$
since $0leq xleq a$.
answered Nov 27 '18 at 1:34
minmax
48518
add a comment |
You know that $Var(X)=int_0^a(x-mu)^2dx=int_0^ax^2p(x)dx-left(int_0^axp(x)dxright)^2$, where
$p(x)geq0$ is the probability distribution and $mu=E(X)$. Hence,
$$
Var(X)leq aint_0^axp(x)dx-mu^2=mu(a-mu)=a^2beta(1-beta)=-a^2(beta^2-beta+1/4-1/4)=-a^2(beta-1/2)^2+a^2/4leq a^2/4
$$
since $0leq xleq a$.
answered Nov 27 '18 at 1:34
minmax
48518
add a comment |
You know that $Var(X)=int_0^a(x-mu)^2dx=int_0^ax^2p(x)dx-left(int_0^axp(x)dxright)^2$, where
$p(x)geq0$ is the probability distribution and $mu=E(X)$. Hence,
$$
Var(X)leq aint_0^axp(x)dx-mu^2=mu(a-mu)=a^2beta(1-beta)=-a^2(beta^2-beta+1/4-1/4)=-a^2(beta-1/2)^2+a^2/4leq a^2/4
$$
since $0leq xleq a$.
answered Nov 27 '18 at 1:34
minmax
48518
You know that $Var(X)=int_0^a(x-mu)^2dx=int_0^ax^2p(x)dx-left(int_0^axp(x)dxright)^2$, where
$p(x)geq0$ is the probability distribution and $mu=E(X)$. Hence,
$$
Var(X)leq aint_0^axp(x)dx-mu^2=mu(a-mu)=a^2beta(1-beta)=-a^2(beta^2-beta+1/4-1/4)=-a^2(beta-1/2)^2+a^2/4leq a^2/4
$$
since $0leq xleq a$.
answered Nov 27 '18 at 1:34
minmax
48518
answered Nov 27 '18 at 1:34
minmax
48518
answered Nov 27 '18 at 1:34
minmax
48518
answered Nov 27 '18 at 1:34
minmax
48518
48518
add a comment |
add a comment |
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is that all the information in the question?
– Siong Thye Goh
Nov 27 '18 at 1:12
@SiongThyeGoh I've edited some typos. And yes, that is all
– Yibei He
Nov 27 '18 at 1:19