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I am trying to solve an equation in the research article. $f(g)$ is the pdf below:

\begin{align} f(g) = \frac{(K+1)e^{-K}}{\bar g}exp\Bigg(-\frac{(K+1)g}{\bar g}\Bigg) I_o\Bigg(\sqrt\frac{4K(K+1)g}{\bar g}\Bigg) \end{align}

$I_o(.)$ is the zeroth order modified Bessel Function of the first kind.

and $\bar g = K+1$.

Here is the complete derivation I have done (following authors hints, since research paper do not solve all the problems):

$$f(g) = e^{-K-g} I_o(\sqrt{4Kg})$$

Now, since $I_n(x)$ has a series representation as:

$$I_n(x) = \sum_{k=0}^\infty \frac{(x/2)^{n+2k}}{(n+k)!\; k!} $$

So for $I_o(.)$, we get

$$ I_o(x) = \sum_{k=0}^\infty \frac{(x/2)^{2k}}{(k!)^2}$$

Hence finally I got $$ f(g) = e^{-K-g} \sum_{k=0}^\infty \frac{(Kg)^k}{(k!)^2} $$

Since $f(g)$ is the pdf, the CDF would be $F_{f(g)}(g)$. Now I do not understand and could not solve beyond this. Author says that the CDF can be simplified by applying 2.321.2 from the following image (reference: Table of Integrals, Series, and Products):

enter image description here

The result that author has yield is :

$$ F_{f(g)}(g) = e^{-K}\sum_{k=0}^\infty \frac{K^k}{k!} \Bigg\{e^{-g}\sum_{m=0}^k(-1)^{2m+1}m! \binom{m}{k} x^{k-m} + k!\Bigg\}$$

I really cannot understand how he came up to this !!! Any idea?

SJa
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1 Answers1

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You have the density

$$f(g) = e^{-K-g} \sum_{k=0}^\infty \frac{K^k g^k}{(k!)^2}\ \text{for}\ g > 0 $$ (the $g>0$ is a detail you left out, but it's important).

For the CDF, you integrate term-by-term: for $x > 0$

$$\eqalign{F(x) &= \sum_{k=0}^\infty \int_0^x e^{-K-g} \frac{K^k g^k}{(k!)^2} \; dg\cr &= e^{-K} \sum_{k=0}^\infty \frac{K^k}{(k!)^2} \sum_{j=0}^k - j! \left. {k \choose j} e^{-g} g^{k-j}\right|_{g=0}^x\cr }$$ Note that ${k\choose j} = \frac{k!}{j!(k-j)!}$ and $\left. e^{-g} g^{k-j}\right|_{g=0}^x = e^{-x} x^{k-j}$ for $j < k$ but $e^{-x} - 1$ for $j=k$. Thus the result becomes

(EDITED)

$$ F(x) = 1 - e^{-K-x} \sum_{k=0}^\infty \frac{K^k}{k!} \sum_{j=0}^k \frac{x^{k-j}}{(k-j)!}$$

I don't know where that $K^{k^2}$ comes from: it must be a typo.

Robert Israel
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  • yes, $K^{k^2}$ is the typo, but still, your answer does not match the answer that I posted for the author's result. I myself am unable to get that result, and yours as well...The values are nearly equal, but not the same.

    But still, many thanks for your time. :)

     – SJa Jan 16 '18 at 22:40
  • Actually, I learned about the typo by reading another paper from the same authors which leads to this paper. But still, the other results are the same in this paper as well. – SJa Jan 17 '18 at 02:21
  • Shouldn't because of $\left. e^{-g} g^{k-j}\right|_{g=0}^x = e^{-x} x^{k-j}$ for $j < k$ but $e^{-x} - 1$ for $j=k$., it should be $[e^{-g}g^{k-j} + e^{-g} - 1]\times A$ ? where A is the rest of the math? I have tried it a thousand times...Could not come to the final conclusion... – SJa Jan 24 '18 at 03:56
  • Extremely sorry for the late response on this, you might thing the matter is solve... ^_^ When you said that $g>0$ is important what do you mean. As I know $g=0$ is a valid condition while considering taking CDF and you also have considered this. – SJa Feb 01 '18 at 23:31
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    A density is defined on the whole real line. This particular one happens to be given by the formula above for $g > 0$ but $0$ for $g \le 0$. The value at $g=0$ is not important, but it is important to not use your formula when $g < 0$. – Robert Israel Feb 02 '18 at 02:29