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Exponent probability density function would be defined as:

$$ f(x)= \begin{cases} \lambda e^{-\lambda x}, & x \ge 0. \\ 0, & x\le 0. \end{cases}$$

Now if i want cumulative distribution function from this i integrate this from $- \infty$ to $\infty$? this should be:

$$F(x)=\int_{-\infty}^x f(s) \, ds=\begin{cases} 1-e^{-\lambda x}, & x> 0. \\ 0, & x\le 0\end{cases}$$

Now there are few things i dont understand. Why we are integrating with different variable than we originally had ? and why we are using $x$ as upper limit ? shouldn't this be $\infty$.

Also if someone could explain all the steps between defining integration and end result. Since i've been trying integrate this by hand but it doesn't seem i would get correct result.

thanks,

Tuki

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  • $\begingroup$ Hint-Use u-substitution $u= - \lambda s$. Also to answer your other question, if the upper bound was positive infinity instead of $x$, the integral would simplify to $1$ instead of of the cumulative distribution function $\endgroup$ – WaveX Oct 21 '17 at 23:34
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The point of a probability density function (lower-case) $f$ of a random varible (capital) $X$ is that for any two numbers $a,b$ with $a<b$ we have $$ \Pr( a\le X\le b) = \int_a^b f(s)\,ds. $$ And the point of a cumulative probability distribution function (capital) $F$ of a random variable $X$ is that for any number $x$ we have $$ F(x) = \Pr(X\le x). $$ Putting these two ideas together we have $$ F(x) = \Pr(X\le x) = \int_{-\infty}^x f(s)\,ds. $$

The variable $s$ goes from $-\infty$ to $x$; it runs through the whole set of numbers that are less than or equal to $x.$ Thus it is not the same thing as $x.$ We could have called it $t,$ and then we have $$ F(x) = \int_{-\infty}^x f(t)\,dt $$ and that's just a valid, provided it's not in a context in which the letter $t$ refers to something else.

Logically it is like the following situation: $$ \sum_{i=1}^3 i^2 = 1^2 + 2^2 + 3^2 = \sum_{j=1}^3 j^2. $$ In the sum $1^2+2^2+3^2$ we don't see anything called $i$ or $j.$ In the first term, $1^2,$ we have $i=1$ or $j=1,$ and in the second term, $2^2,$ we have $i=2$ or $j=2.$ We can call that "bound variable" either $i$ or $j$ or $k$ or something else.

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  • $\begingroup$ Thanks for providing this. This clarified the upper bound variable i didn't first understand and now i do. It makes sense. When integrating within bounds it's usually (atleast this is how i remember this) $$F(x)= \int_{-\infty}^{x}f(s)ds=\lim_{n\rightarrow (-\infty)} \int_{n}^{x}f(s)ds=\lim_{n\rightarrow (-\infty)}\left[\begin{matrix} \end{matrix} F(x)-F(n) \right] $$ But i end up with resulting looking like this. $$ F(x)= \int_{-\infty}^{x}\lambda e^{-\lambda s}ds=\lim_{n\rightarrow(-\infty)}(-e^{-\lambda x})-(-e^{-\lambda n}) $$ now how do you get from here to the end ? $\endgroup$ – Tuki Oct 22 '17 at 10:25
  • $\begingroup$ @Tuki : Note that $$f(s) = \begin{cases} \lambda e^{-\lambda s} & \text{if } s\ge 0, \\ 0 & \text{if } s <0. \end{cases}$$ Therefore $$ \int_{-\infty}^x f(s)\,ds = \int_0^x \lambda e^{-\lambda s} \, ds \text{ if } x\ge 0. $$ This becomes $ -e^{-\lambda x} - ( - e^0) = 1 - e^{-\lambda x}. \qquad$ $\endgroup$ – Michael Hardy Oct 22 '17 at 18:50

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