The cumulative distribution function \Phi : \mathbb {R}\to \mathbb {R} of the standard normal \mathcal{N}(0,1) can be written as
\Phi (z) = \int _{A}^{B(z)} \frac{1}{\sqrt{2 \pi }} e^{C(x)} \, dx
where B(z) is a function of z and C(x) is a function of x. Write down the integration limits A,B(z),\, as well as the function \, C(x) in the integrand.
Enter inf for \infty.
A=\quad
unanswered
B=\quad
unanswered
C=\quad
A=-\infty
B(z)=z
C(x)=-\frac{x^2}{2}
To determine the integration limits A and B(z), as well as the function C(x) in the integrand, we need to consider the cumulative distribution function (CDF) of the standard normal distribution.
The standard normal distribution is defined by its probability density function (PDF):
f(x) = \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}x^2}
The CDF, denoted as \Phi(z), is given by integrating the PDF from negative infinity to z:
\Phi(z) = \int_{-\infty}^{z} \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}x^2} dx
To rewrite the CDF in the form mentioned, we need to determine the integration limits A and B(z) and the function C(x) in the integrand.
A: Since the integration is from negative infinity (-\infty), the lower limit A is negative infinity.
A = -\infty
B(z): The upper limit of the integration B(z) is given by z, as mentioned.
B = z
C(x): The function C(x) in the integrand is equal to -\frac{1}{2}x^2.
C(x) = -\frac{1}{2}x^2
Therefore, the integration limits and the function C(x) in the integrand are:
A = -\infty
B = z
C(x) = -\frac{1}{2}x^2