Lemma for Boundaries of standard normal distribution

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I have a problem in understanding the proof of my professor. I hope you can tell me how to understand his proof and alternatively I would like to show how I would proof this.



I hope you can take a look into this and help me out.



My profs theorem and proof:



For x > 0 the following equation holds.
beginequation
frac1sqrt2pixe^frac-x^22(1-frac1x^2)leq 1- Phi(x) leq frac1sqrt2pixe^frac-x^22
endequation



Proof by partial integration:
beginequation
int_x^inftyfrac1uue^frac-u^22du = frac1xe^x^2/2-int_x^inftyfrac1u^2e^frac-u^22du = (frac1x-frac1x^3)e^-x^2/2+3int_x^inftyfrac1u^4e^frac-u^22du
endequation



For every $delta>0$ is now
beginequation
1-Phi(x) geq e^-(1+delta)x^2/2
endequation
if x is large enough $square$



I have no idea what this is showing.
Now my naive approach:



I would define at first some functions.
$f(x) :=frac1sqrt2pie^frac-x^22; Phi(x):= int_-infty^x f(t) dt;g(x):=1-frac1x^2$ and substitute the inequality.
beginequation
fracf(x)xgleq 1 - int_-infty^x f(t) dt leq fracf(x)x:= A leq B leq C
endequation
The case $A<C$ is trivial and for the case $Bleq C$ and $Aleq B$ I would like to use the fundamental theorem of calculus, which gives me the following:
beginequation
fracdBdx leq fracdCdx iff f(x) leq (1+frac1x^2) f(x)
endequation
beginequation
fracdAdx leq fracdBdxiff (1-frac3x^4) f(x) leq f(x)
endequation



I would like to know if my approach is right. Thank you very much :)




probability inequality




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  • It looks OK. It is a little confusing, since the derivatives all have minus signs!
    – herb steinberg
    Jul 30 at 17:12














up vote
0
down vote

favorite












I have a problem in understanding the proof of my professor. I hope you can tell me how to understand his proof and alternatively I would like to show how I would proof this.



I hope you can take a look into this and help me out.



My profs theorem and proof:



For x > 0 the following equation holds.
beginequation
frac1sqrt2pixe^frac-x^22(1-frac1x^2)leq 1- Phi(x) leq frac1sqrt2pixe^frac-x^22
endequation



Proof by partial integration:
beginequation
int_x^inftyfrac1uue^frac-u^22du = frac1xe^x^2/2-int_x^inftyfrac1u^2e^frac-u^22du = (frac1x-frac1x^3)e^-x^2/2+3int_x^inftyfrac1u^4e^frac-u^22du
endequation



For every $delta>0$ is now
beginequation
1-Phi(x) geq e^-(1+delta)x^2/2
endequation
if x is large enough $square$



I have no idea what this is showing.
Now my naive approach:



I would define at first some functions.
$f(x) :=frac1sqrt2pie^frac-x^22; Phi(x):= int_-infty^x f(t) dt;g(x):=1-frac1x^2$ and substitute the inequality.
beginequation
fracf(x)xgleq 1 - int_-infty^x f(t) dt leq fracf(x)x:= A leq B leq C
endequation
The case $A<C$ is trivial and for the case $Bleq C$ and $Aleq B$ I would like to use the fundamental theorem of calculus, which gives me the following:
beginequation
fracdBdx leq fracdCdx iff f(x) leq (1+frac1x^2) f(x)
endequation
beginequation
fracdAdx leq fracdBdxiff (1-frac3x^4) f(x) leq f(x)
endequation



I would like to know if my approach is right. Thank you very much :)




probability inequality




share|cite|improve this question



















  • It looks OK. It is a little confusing, since the derivatives all have minus signs!
    – herb steinberg
    Jul 30 at 17:12












up vote
0
down vote

favorite









up vote
0
down vote

favorite











I have a problem in understanding the proof of my professor. I hope you can tell me how to understand his proof and alternatively I would like to show how I would proof this.



I hope you can take a look into this and help me out.



My profs theorem and proof:



For x > 0 the following equation holds.
beginequation
frac1sqrt2pixe^frac-x^22(1-frac1x^2)leq 1- Phi(x) leq frac1sqrt2pixe^frac-x^22
endequation



Proof by partial integration:
beginequation
int_x^inftyfrac1uue^frac-u^22du = frac1xe^x^2/2-int_x^inftyfrac1u^2e^frac-u^22du = (frac1x-frac1x^3)e^-x^2/2+3int_x^inftyfrac1u^4e^frac-u^22du
endequation



For every $delta>0$ is now
beginequation
1-Phi(x) geq e^-(1+delta)x^2/2
endequation
if x is large enough $square$



I have no idea what this is showing.
Now my naive approach:



I would define at first some functions.
$f(x) :=frac1sqrt2pie^frac-x^22; Phi(x):= int_-infty^x f(t) dt;g(x):=1-frac1x^2$ and substitute the inequality.
beginequation
fracf(x)xgleq 1 - int_-infty^x f(t) dt leq fracf(x)x:= A leq B leq C
endequation
The case $A<C$ is trivial and for the case $Bleq C$ and $Aleq B$ I would like to use the fundamental theorem of calculus, which gives me the following:
beginequation
fracdBdx leq fracdCdx iff f(x) leq (1+frac1x^2) f(x)
endequation
beginequation
fracdAdx leq fracdBdxiff (1-frac3x^4) f(x) leq f(x)
endequation



I would like to know if my approach is right. Thank you very much :)




probability inequality




share|cite|improve this question











I have a problem in understanding the proof of my professor. I hope you can tell me how to understand his proof and alternatively I would like to show how I would proof this.



I hope you can take a look into this and help me out.



My profs theorem and proof:



For x > 0 the following equation holds.
beginequation
frac1sqrt2pixe^frac-x^22(1-frac1x^2)leq 1- Phi(x) leq frac1sqrt2pixe^frac-x^22
endequation



Proof by partial integration:
beginequation
int_x^inftyfrac1uue^frac-u^22du = frac1xe^x^2/2-int_x^inftyfrac1u^2e^frac-u^22du = (frac1x-frac1x^3)e^-x^2/2+3int_x^inftyfrac1u^4e^frac-u^22du
endequation



For every $delta>0$ is now
beginequation
1-Phi(x) geq e^-(1+delta)x^2/2
endequation
if x is large enough $square$



I have no idea what this is showing.
Now my naive approach:



I would define at first some functions.
$f(x) :=frac1sqrt2pie^frac-x^22; Phi(x):= int_-infty^x f(t) dt;g(x):=1-frac1x^2$ and substitute the inequality.
beginequation
fracf(x)xgleq 1 - int_-infty^x f(t) dt leq fracf(x)x:= A leq B leq C
endequation
The case $A<C$ is trivial and for the case $Bleq C$ and $Aleq B$ I would like to use the fundamental theorem of calculus, which gives me the following:
beginequation
fracdBdx leq fracdCdx iff f(x) leq (1+frac1x^2) f(x)
endequation
beginequation
fracdAdx leq fracdBdxiff (1-frac3x^4) f(x) leq f(x)
endequation



I would like to know if my approach is right. Thank you very much :)





probability inequality





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share|cite|improve this question




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asked Jul 30 at 11:53









Markus Krumpl

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  • It looks OK. It is a little confusing, since the derivatives all have minus signs!
    – herb steinberg
    Jul 30 at 17:12
















  • It looks OK. It is a little confusing, since the derivatives all have minus signs!
    – herb steinberg
    Jul 30 at 17:12















It looks OK. It is a little confusing, since the derivatives all have minus signs!
– herb steinberg
Jul 30 at 17:12




It looks OK. It is a little confusing, since the derivatives all have minus signs!
– herb steinberg
Jul 30 at 17:12















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