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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 $X_i$? –  joriki Aug 1 '12 at 11:43
    
Yes, will correct it in the OP. –  Omri Aug 1 '12 at 11:49
    
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. –  Did Aug 1 '12 at 12:13
    
OK, thanks, the pgf works only for $a_i=1$ using a little trick (and useful only if each $X_i$ has a $p_i$).. –  Omri Aug 1 '12 at 12:28
    
The OP has now been restated. –  Omri Aug 1 '12 at 14:16
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1 Answer

up vote 1 down vote accepted

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? –  Omri Aug 5 '12 at 6:23
    
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? –  Did Aug 5 '12 at 6:30
    
The reason I want to see all the stages is that my model also has $(X_i)$ as discrete $i.i.id$ RVs with $P(0)=(1-p_i)^2$, $P(1)=2p_i(1-p_i)$, $P(2)=p_i^2$. –  Omri Aug 5 '12 at 9:00
    
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? –  Did Aug 6 '12 at 14:29
    
If I knew the exact stages myself I would not have bothered anyone to provide the solution. Thanks anyway. –  Omri Aug 8 '12 at 7:54
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