May 2011 Q.1 IAI

Hello Everyone

If Z ~ N(0,1), then for any a > 0 show that

P[Z > a] is less than or equal to 1/2*exp{-1/2a^2)

The first thing that came to my mind is that we have use some kind of relation between standard normal and exponential distribution but could not get any ground on it

Can anyone help me?

Regards,
Kartik.

 

This question is quite simple if you know how to approach it..

By definition P(Z>a) = Integral(a,+infinity) 1/sqrt(2pi) * exp (-1/2*z^2) dz

Now using the standard formulae for normal truncated moments this is equal to

= -[phi(infinity) - phi(a)]

= - 0 + 1/sqrt(2*pi) * exp(-1/2*a^2)

<= 1/2*exp(-1/2*a^2)

QED.

 

Ummm .... Well ..... I have completed only first 9 chapters from CT3 course. Does this proof lie out of those 9 chapters? (Because I dont know what are normal truncated moments)

 

Theyre somwhere in the formula book (They are proved formally in CT6) but they don't really make use of any mathematics that you shouldn't be familar with so i wouldn't worry too much.

 

try doing the integral by transforming the integral in such a way as to look like a full standard normal integral...u will get the answer..not sure whether formulae for evaluating truncated integral was introduced in CT 3. As I remember, it was not..

 

Incorrect

I think the derivation below by DevonMatthews is incorrect.

To check, just set a=0.
LHS = P(Z>a) = 0.5; RHS = phi(a) = 1/sqrt(2*pi) * exp(-1/2*0^2) = 1/sqrt(2*pi). This can't be equal then.

Actually I think to prove this, we don't need to know/use the truncated moments.

First set x = z – a in the formula of P(Z>a). Take the constant terms outside the integral. Using the fact that exp(-x.a) <= 1, note that the resultant integral value is sqrt(pi/2).

You then get LHS <= 1/2*exp(-1/2*a^2)

 

Yeah thanks. This sounds quite reasonable.

But in an exam how do you get an intuition that you have to use this method?

Honestly, I racked my brains for too long before coming to this forum. Is there some kind of intuitive method to solve such problems?

Or am I just not smart enough?

 

i agree it requires a bit of intuition ... actually you know how i guessed this .. the RHS had a Exp(-a^2) term .. usually such terms can only come out of the normal integral ... and a transforamtion like x = z - a works most of the times .. something i learned to apply during my senior school years while doing integration ...

 

Hey bapan,

I could not get it right though I thought I understood your method. Can you just elaborate on your proof a little.

Thanks.

 

Set z = x - a.
Replace that in the integral.
LHS = exp (-1/2*a^2) * Integral(0,+infinity) 1/sqrt(2pi) * exp (-1/2*x^2) * exp(-x.a) dx
<= exp (-1/2*a^2) * Integral(0,+infinity) 1/sqrt(2pi) * exp (-1/2*x^2) dx [as exp(-x.a) <= 1]
= 0.5 * exp (-1/2*a^2) [as the above integral value is sqrt(pi/2)]

Trust this helps

 

Devon, this is intriguing (can't say that I remember much about truncated moments).
Care to elaborate?
Did you mean capital or common phi? 1st to 2nd line works if capital but then 2nd to 3rd seems like common.


Kartik, Bapan's method is probably best but alternatively I did it by writing the integral, integrating by parts and discarding the extra positive integral of x^2/sqrt(2pi) e^((-x^2)/2)), then noting that 1/sqrt(2pi)<1/2