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I am solving a problem where $X$ is an exponential random variable and $\lambda=\frac{1}{10}$. I need to find the CDF of $X$ and have that $\int_0^\infty \frac{e^\frac{-x}{10}}{10}$ turns out to be $1$, however the answer is $1-e^\frac{-x}{10}$. Where did I go wrong?

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The cumulative distribution function, $F_X$, is a function; its value at the point $a\ge 0$ is $$F_X(a)=P[X\le a]=\int_0^a{\textstyle {1\over10}} e^{-x/10}\,dx.$$ Note, also, that $F_X(a)=0$ for $a<0$.

Of course, use $x$ if you like for the independent variable.

After doing the integration, you should find: $$ F_X(x)=\cases{1-e^{-x/10},&$x\ge0$\cr 0,&$x<0$ }. $$

What you found was "$F(\infty)$", which is of course 1.

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So (completing the indefinite integral) we have $\int\frac{e^\frac{-x}{10}}{10}= -e^\frac{-x}{10}$ and for any value $x$ less than $\infty$, $F(x)=1-e^\frac{-x}{10}$. Is that the correct reasoning? –  johnnymath Mar 19 '12 at 23:19
    
@johnnymath Not exactly. You compute a definite integral for the CDF. $F(a)$ is the probability that $X$ is at most $a$. So you integrate the density from $0$ (here, since the density is 0 for $x<0$) to $a$: $F(a)=\int_0^a {1\over10} e^{-x/10}\,dx$. So, you would have $F(a)= (-e^{-x/10})\bigl|_0^a=...$. –  David Mitra Mar 19 '12 at 23:25
    
@johnnymath Your final answer in the comment is correct for $x\ge0$. But if $x<0$, then $F(x)=0$. The cdf is usually defined for all $x$. Since your $X$ does not take negative values, for $x<0$, $F(x)=P[X\le x]=0$. –  David Mitra Mar 19 '12 at 23:29
    
While it is for a different pdf, you might want to read this and the comments thereafter. –  Dilip Sarwate Mar 20 '12 at 2:05
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