Need help with figuring out this limit

[DanCali]

I am stuck on this problem. Any help would be greatly appreciated!


lim as T-> infinity of:

Numerator: 1/sqrt(2*pi)*(s/sqrt(T))*exp(-0.5*(a/s)^2*T)

denominator: N(-(a/s)*sqrt(T))

a is assumed positive. s is assumed positive (standard deviation of normally distributed variable with mean zero)

N is the cumulative distribution function of a standard normal variable.

Since both numerator and denominator converge to zero, this probably required l'hospital's rule. However, it's probably harder than it looks.

 

[Deleted member 4993]

I am stuck on this problem. Any help would be greatly appreciated!


lim as T-> infinity of:

Numerator: 1/sqrt(2*pi)*(s/sqrt(t))*exp(-0.5*(a/s)^2*T ---- Where is the closure of that paranthesis?

denominator: N(-(a/s)*sqrt(T))

a is assumed positive. s is assumed positive (standard deviation of normally distributed variable with mean zero)

N is the cumulative distribution function of a standard normal variable.

Since both numerator and denominator converge to zero, this probably required l'hospital's rule. However, it's probably harder than it looks.

Your expression contain "T" and "t" - are those same?

If yes - correct your post and tell us so.

If no - tell us so.

 

[DanCali]

Your expression contain "T" and "t" - are those same?

If yes - correct your post and tell us so.

If no - tell us so.

Yes, they are the same. I edited the original post. Thanks!

 

[Deleted member 4993]

I am stuck on this problem. Any help would be greatly appreciated!


lim as T-> infinity of:

Numerator: 1/sqrt(2*pi)*(s/sqrt(T))*exp(-0.5*(a/s)^2*T)

denominator: N(-(a/s)*sqrt(T))

a is assumed positive. s is assumed positive (standard deviation of normally distributed variable with mean zero)

N is the cumulative distribution function of a standard normal variable.

Since both numerator and denominator converge to zero, this probably required l'hospital's rule. However, it's probably harder than it looks.

How did you get that?

\(\displaystyle \lim_{T \to \infty}[N * (-\frac{a}{s}) * \sqrt{T}] \ = \ \infty\)

 

[DanCali]

How did you get that?

\(\displaystyle \lim_{T \to \infty}[N * (-\frac{a}{s}) * \sqrt{T}] \ = \ \infty\)

OK - i think I figured a way to clarify the expression:

\(\displaystyle $\lim_{t->\infty }\frac{\frac{1}{\sqrt{2\pi }}\left( \frac{\sigma _{\epsilon
}}{\sqrt{T}}\right) e^{-0.5\left( \frac{a}{\sigma _{\epsilon }}\right) ^{2}T}%
}{\Phi \left( -\left( \frac{a}{\sigma _{\epsilon }}\right) \sqrt{T}\right) }$\)

The denominator is the cdf of standard normal. i.e.:

\(\displaystyle \Phi \left( -\left( \frac{a}{\sigma _{\epsilon }}\right) \sqrt{T}\right) =%
\frac{1}{\sqrt{2\pi }}\int_{-\infty }^{-aT}\frac{e^{-0.5\left( \frac{%
\epsilon }{\sigma _{\epsilon }\sqrt{T}}\right) ^{2}}}{\sigma _{\epsilon }%
\sqrt{T}}d\epsilon
\)
where epsilon has mean zero and standard deviation sigma*sqrt(T)


a is a positive constant, so the denominator goes to zero as T goes to infinity.