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# Standard Normal Distribution Function

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Let Phi(x) be the cumulative distribution function in the standard normal distribution.

a) Use L'Hopital's rule to show that the limit as x goes to - infinity of xPhi(x) equals 0.

b) Show that sign [(d/dx)(Phi(x)/Phi'(x))] = sign (Phi'(x) + Phi(x)x).

c) Use (a) and (b) to show that Phi(x)/Phi'(x) is increasing in x over R.

##### Solution Summary

For the cumulative distribution function of the standard normal distribution Phi(x), the integral from -infinity to x of 1/sqrt(2 pi)e^(-t^2/2), it is shown that the ratio Phi(x)/Phi'(x) is an increasing function of x.

##### Solution Preview

For part (a),

Remember that the L'Hopital's rule apply to limits of the form

lim_(x->x_0) f(x)/g(x)

in one of the following cases

(1)---- lim_{x->x_0}f(x)=0 and lim_{x->x_0}g(x)=0
(2)---- lim_{x->x_0}f(x)=plus or minus infinity and lim_{x->x_0}g(x)=plus or minus infinity,

and the rule is

lim_(x->x_0) f(x)/g(x) =lim_(x->x_0) f'(x)/g'(x).

In the case of the question, we note that the limit of the function Phi(x) as x->-infinity is zero

lim_{x->-infinity}Phi(x)=0,

since the lower limit of the integral is -infinity.

Now, the limit is not of the form (1) nor (2) above but we manipulate it:

lim_{x->-infinity} x Phi(x)=lim_{x->-infinity} Phi(x)/(1/x)

so now the numerator, Phi(x), and denominator, 1/x, both have limit 0 as x->-infinity, and so we are in case (1) of L'Hopital's rule. Applying L'Hopital's rule

lim_{x->-infinity} Phi(x)/(1/x)=lim_{x->-infinity} Phi'(x)/(1/x)'
=lim_{x->-infinity} 1/sqrt(2 pi) e^(-x^2/2)/(-1/x^2),

where we used the fundamental theorem of calculus to calculate Phi'(x)=1/sqrt(2 pi) e^(-x^2/2), and the derivative of 1/x is -1/x^2. With this the limit is

lim_{x->-infinity} 1/sqrt(2 pi) e^(-x^2/2)/(-1/x^2)=lim_{x->-infinity} -1/sqrt(2 pi) x^2e^(-x^2/2)

This last limit can be done in ...

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