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Say $(X_i)$ is a sequence of $n$ $i.i.d$ Bernoulli random variables, each with parameter $p_i$. Define the following for all real $a_i>0$, $$ S_{n} =\frac{1}{n} \sum _{i=1}^{n}a_iX_i $$ EDIT: In the special case of $a_i=1$, we may use probability generating functions to arrive at the probability mass function of $S_n$,

$$P_{S_{n} } (k)=\frac{1}{(nk)!} \frac{d^{(nk)} }{dx}\prod_{i=1}^{n} \left(1-p_i+p_i\cdot x \right) \; \left|\; x=0\right.$$

for $k=0,1/n,2/n,...,1$. Is there any other way to get an expression for the probability mass function of $S_n$ that will be defined for all real $a_i>0$?

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    Presumably $X$ in the sum is meant to be $X_i$?2012-08-01
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    Yes, will correct it in the OP.2012-08-01
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    Nothing guarantees that $S_n$ is integer-valued when the $a_i$ are not integer. Since generating functions are well defined only for (nonnegative) integer-valued random variables, unsurprisingly one cannot use them in this case. The solution is to turn to Fourier transforms, known to probabilists as characteristic functions.2012-08-01
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    OK, thanks, the pgf works only for $a_i=1$ using a little trick (and useful only if each $X_i$ has a $p_i$)..2012-08-01
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    The OP has now been restated.2012-08-01
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    What part of the restated question is not answered by my previous comment?2012-08-02
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    You provided a good lead, thanks, but maybe someone could provide the solution.2012-08-02

1 Answers 1

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Hint: Use the fact that, for every integer valued random variable $S$ and every integer $x$, $$ \mathrm P(S=x)=\int_0^{1}\mathrm E(\mathrm e^{2\pi\mathrm itS})\cdot\mathrm e^{-2\pi\mathrm itx}\,\mathrm dt, $$ and the fact that, in the present case, $$ \mathrm E(\mathrm e^{2\pi\mathrm itS_n})=\prod_{k=1}^n(1-p_k+p_k\mathrm e^{2\pi\mathrm ita_k/n}). $$ Edit: The second identity above is a consequence of the definition of $\mathrm E(\mathrm e^{2\pi\mathrm itS_n})$ and of the joint distribution of the random variables $(X_k)_{1\leqslant k\leqslant n}$.

The first identity is an application of the general principle that integrating a discrete sum of complex exponentials against the conjugate of a complex exponential extracts the coefficient of the corresponding exponential from the sum. Namely, for every integers $x$ and $y$, $$ \int_0^{1}\mathrm e^{2\pi\mathrm ity}\cdot\mathrm e^{-2\pi\mathrm itx}\,\mathrm dt=[x=y], $$ hence, for every distinct integers $x_k$ and every coefficients $p_k$, $$ p_\ell=\int_0^{1}\left(\sum_kp_k\mathrm e^{2\pi\mathrm itx_k}\right)\cdot\mathrm e^{-2\pi\mathrm itx_\ell}\,\mathrm dt. $$ Applying this to the integer valued random variable $S$ such that $p_k=\mathrm P(S=x_k)$ yields the first formula above.

When $S$ is not integer valued, use the fact that, for every real numbers $x$ and $y$, $$ \lim_{N\to\infty}\int_0^1\mathrm e^{N\mathrm ity}\cdot\mathrm e^{-N\mathrm itx}\,\mathrm dt=[x=y], $$ hence, for every discrete random variable $S$, $$ \mathrm P(S=x)=\lim_{N\to\infty}\int_0^1\mathrm E(\mathrm e^{N\mathrm itS})\cdot\mathrm e^{-N\mathrm itx}\,\mathrm dt. $$

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    Thanks much, @did. Is there any way you could elaborate on the preceding stages of this, so I could see how it was arrived at?2012-08-05
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    Is there any way you could elaborate on the exact stages of this solution causing you problems, so I could see how it can be completed?2012-08-05
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    The reason I want to see all the stages is that my model also has $(X_i)$ as discrete $i.i.id$ RVs with $P(0)=(1-p_i)^2$, $P(1)=2p_i(1-p_i)$, $P(2)=p_i^2$.2012-08-05
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    You explained your reasons, this was not my suggestion. So, once again: *is there any way you could elaborate on the exact stages of this solution causing you problems, so I could see how it can be completed?*2012-08-06
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    If I knew the exact stages myself I would not have bothered anyone to provide the solution. Thanks anyway.2012-08-08
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    Once again this is unrelated. Let me try: (1.) Is the first identity a problem? If yes, can you describe it? (2.) Is the second identity a problem? If yes, can you describe it?2012-08-08
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    Both identities are not clear (to me) how they were reached. If you could include earlier stages I might be able to modify the solution to apply to the discrete RV I described above.2012-08-09
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    See Edit. // In the specific case when $P(X_i=0)=(1-p_i)^2$, $P(X_i=1)=2p_i(1-p_i)$ and $P(X_i=2)=p_i^2$, one may want to realize each $X_i$ as the sum of two i.i.d. Bernoulli random variables with parameter $p_i$. Then the sum of $n$ independent random variables $X_i$ is the sum of $2n$ independent Bernoulli random variables hence one is back to the elementary case.2012-08-09
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    Thanks much, @did. I am processing it...2012-08-09
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    OK... $ $ $ $ $ $2012-08-09