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There is some details i don't understand in my book, here goes;

Let $X \sim N(\mu,\sigma^2)$ and $Z\sim N(0,1)$

we know that:

$$F_X(x) = \int\limits_{-\infty}^{x} \frac{1}{\sigma \sqrt{2\pi}}e^{-\frac{(t-\mu)^2}{2\sigma^2}} \, dt $$

(this is the part i don't understand , i could understand some if the substitution is $z = (t-\mu$)/ \sigma , but then i wouldn't understand how to get the upper limit of the integral , that is : $(x - \mu)/\sigma$ )

with variable substitution $z = (x - \mu)/\sigma $ we can get:

$$F_X(x)= \int\limits_{-\infty}^{(x-\mu)/\sigma} \frac{1}{\sqrt{2\pi}}e^{-z^2/2} dz = \Phi\left(\frac{x-\mu}{\sigma}\right)$$

remark: $\Phi(z) = \int\limits_{-\infty}^{z}\frac{1}{\sqrt{2\pi}}e^{-t^2/2} dt$

can someone fill in with more details so that i could understand what is going on here

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  • $\begingroup$ The intended substitution is $$z = \frac{t-\mu}{\sigma},$$ not $z = (x-\mu)/\sigma$. Then $dz = \frac{1}{\sigma} \, dt$, and the upper limit of integration clearly becomes $(x-\mu)/\sigma$. $\endgroup$ – heropup Feb 1 '14 at 0:54
  • $\begingroup$ @heropup "making the computations" from left to right in the second integral , with substitution $z = \frac{x- \mu}{\sigma}$ where $x = t$. then the upper limit of integration should be $z\sigma + \mu = x$ ? . $\endgroup$ – Danny Feb 1 '14 at 1:25
  • $\begingroup$ When the integration is with respect to $t$, then the limits are $t \in (-\infty, x]$. Then if $z = (t-\mu)/\sigma$, it follows that $z \in (-\infty, (x-\mu)/\sigma]$. $\endgroup$ – heropup Feb 1 '14 at 1:50
  • $\begingroup$ As Michael Hardy says $z=\frac{t-\mu}{\sigma}$. If $t$ goes up to $x$ then $z$ would go up to $\frac{x-\mu}{\sigma}$. $\endgroup$ – robjohn Feb 1 '14 at 2:11
  • $\begingroup$ thanks guys i got it now!! $\endgroup$ – Danny Feb 1 '14 at 2:45
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\begin{align} z & = \frac{t-\mu}{\sigma} \\[10pt] dz & = \frac{dt}{\sigma} \end{align}

When $t=x$ then $z=\text{what?}$

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