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Let the distribution function of $X$ for $x>0$ be:

$$F(x) = 1 - \sum_0^3 \frac{x^ke^{-x}}{k!}$$

what is the density function of $X$ for $x > 0$?


This is what I'm thinking:

$$ \frac{x^0e^{-x}}{1} + \frac{x^1e^{-x}}{1} + \frac{x^2e^{-x}}{2} + \frac{x^3e^{-x}}{6} $$

and this looks like the beginning of a series expansion... and I know one thing we can do with series is try to differentiate or integrate.

And since we are going from cdf to pdf I think that means we want to differentiate (so maybe I didnt need to actually plug those numbers in above)

So: $$ \frac d{dx}(1-\sum_0^3 \frac{x^ke^{-x}}{k!}) = $$ I don't know but here goes nothin...:$$-( kx^{k-1}e^{-x}-x^ke^{-x} ) =$$ I think thats the derivative of the numerator anyway. Now I'm seeing k! as a constant no matter what. So that denominator can just be left there til after? I guess the summation can't just disappear for no good reason either so: $$-\sum_0^3 \frac {kx^{k-1}e^{-x}-x^ke^{-x}}{k!} $$

OK so my reasoning about the why i THOUGHT it is discrete is because I saw the summation and thought 'ooh, discrete' but it's actually increasing the k by 1, not the x, and the problem states x>0.

So lemme try this again: $$-e^{-x}\sum_0^3 \frac {kx^{k-1}-x^k}{k!} = $$ $$e^{-x}\sum_0^3 \frac {x^k-kx^{k-1}}{k!} = $$

so now I can use my summation knowledge!...

Ooh wow, so this is interesting, the first 4 terms: $$ (1) + (x-1) + (\frac12 x^2-x) + (\frac{x^3}6-\frac12 x^2)$$

so each successive term is negating the previous 2nd highest power's term.

Therefore: $$ (\frac16)e^{-x}x^3 $$

phew! But super interesting! What is this sort of series called? I like its style (I haven't done much with series)

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    $\begingroup$ At first I noticed it is discrete. Also, the sum = 1 which makes sense being that it is a cdf. Accidentally pressed 'enter'.. im still writing $\endgroup$
    – Adam
    Sep 19, 2014 at 3:59
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    $\begingroup$ and distribution functions is saying the probability added up to this point is... ARGH why does alt+enter not just put me to a new line. Sorry about the multiple comments, I'll continue writing in the next one $\endgroup$
    – Adam
    Sep 19, 2014 at 4:02
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    $\begingroup$ shift+enter lets you write in a new line. Note that the new lines don't appear in the final comment, however. $\endgroup$ Sep 19, 2014 at 4:14
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    $\begingroup$ Are you sure this is discrete? $\endgroup$ Sep 19, 2014 at 4:16
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    $\begingroup$ As for the "$1-$": note that any cdf defined over $x>0$ has to have a limit of $1$ as $x \to \infty$ $\endgroup$ Sep 19, 2014 at 4:20

1 Answer 1

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For every nonnegative integer $n$, the density $f$ defined on $x\gt0$ by $f(x)=\frac1{n!}x^n\mathrm e^{-x}$ is the PDF of the gamma distribution $(n+1,1)$. Your case is when $n=3$. The parameter $n$ could be any real number $n\gt-1$ provided $n!$ is replaced by $\Gamma(n+1)$.

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