Sorry for the vague title--I'm not sure how to summarize this succinctly.

We have a random variable $Z$ parameterized by a vector $u\in[0,1]^n$ given by

$$ Z(u)=\sum_{i=1}^n\sum_{j=1}^{Y_i}(X_{ij}-\alpha) $$

where $\alpha\in\mathbb{R}$ is a constant, $X_{ij}$ are iid with $X_{ij}\sim\text{Bernoulli}(p)$ for fixed $p\in[0,1]$, and $Y_i\sim\text{Poisson}(\lambda{u_i})$ for some $\lambda>0$.

Now, if $\alpha=0$, then we have that

$$ Z(u)\sim\text{Poisson}\bigg(p\lambda\sum_{i=1}^nu_i\bigg). $$

With this fact, I can then show that for any $u\in[0,1]^n$, the following identity holds

$$ \mathbb{P}\big[Z(u)\leqslant0\big]=\mathbb{P}\big[Z(\overline{u})\leqslant0\big] \quad(**) $$ where $$\overline{u}=\bigg(\frac{1}{n}\sum_{i=1}^nu_i\bigg)\mathbf{1} $$ and $\mathbf{1}\in\mathbb{R}^n$ is the vector of all ones.

My question is, does this property $(**)$ hold if $\alpha\neq0$? In this case it doesn't seem like $Z$ follows any known distribution, so I don't know how to calculate the probabilities.


I think I figured it out using characteristic functions.

We wish to compute the characteristic function of $Z$. So let's break it down using some properties of characteristic functions. First, let's compute the characteristic function of $X_{jk}-\alpha$. This is given by

$$\varphi_{X-\alpha}(t)=e^{-\iota\alpha{t}}\big(1-p+pe^{\iota{t}}\big) $$

where $\iota=\sqrt{-1}$. Then, according to Wikipedia, the characteristic function of


is given by

$$ \varphi_{W_j}(t)=\exp\big[\lambda{u_j}(\varphi_{X-\alpha}(t)-1)\big]$$

which gives

$$ \varphi_{W_j}(t)=\exp\big[\lambda{u_j}(e^{-\iota\alpha{t}}\big(1-p+pe^{\iota{t}}\big)-1)\big]$$

Finally, assuming that the $Y_j$ are independent for all $j$, the characteristic function of

$$ Z=\sum_{j=1}^nW_j $$

is given by

$$ \varphi_Z(t)=\varphi_Z(t;u)=\prod_{j=1}^n\varphi_{W_j}(t)=\exp\bigg[\lambda\big(e^{-\iota\alpha{t}}(1-p+pe^{\iota{t}})-1\big)\sum_{j=1}^nu_j\bigg],$$

or, more compactly,


where $\xi(t)=\xi(t;p,\alpha,\lambda)=\lambda(e^{-\iota\alpha{t}}(1-p+pe^{\iota{t}})-1)$.

Now, fix $u\in[0,1]^n$ and $t\in\mathbb{R}$, and define $\overline{u}$ by

$$ \overline{u}_k=\frac{1}{n}\sum_{j=1}^nu_j$$ for $k=1,\dots,n$. Then

$$ \varphi_Z(t;\overline{u})=\exp\bigg[\xi(t)\sum_{k=1}^n\overline{u}_k\bigg]=\exp\bigg[\xi(t)\sum_{k=1}^n\frac{1}{n}\sum_{j=1}^nu_j\bigg]=\exp\bigg[\xi(t)\sum_{j=1}^nu_j\bigg]=\varphi_Z(t;u).$$

Hence $Z(u)$ and $Z(\overline{u})$ have the same characteristic function $\Rightarrow$ they have the same CDF. $\square$

  • $\begingroup$ plus one for self-answer and sharing the knowledge. $\endgroup$ – Lee David Chung Lin Jun 14 '18 at 23:16

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