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Let $X_1, \dots, X_c$ be independent uniform random variables on $[0,1]$. We know the number of random variables whose value will fall in the range $[0,1]$ will always be exactly $c$. Now define $Z_i = X_i + Y_i$ where $Y_i$ are independent random variables that are exponentially distribution. Let $C_Z$ be the number of random variables $Z_i$ whose value falls in the range $[0,1]$. I am interested in the distribution of $C_Z$.

If we set $c=100$ and $\lambda = 1$ then, by simulation the pdf of $C_Z$ looks like:

enter image description here

I fitted a normal distribution to the data and have drawn that on top of the histogram created by simulation. This is just to show the data appears to be approximately normal.

It seems that $C_Z$ has a shifted binomial distribution. Is that right and if so, why?


Update

@E-A points out that the pdf is binomial with parameters (p, c) with $p = P(X_i + Y_i \leq 1)$. But what is $P(X_i + Y_i \leq 1)$? By simulation again I plotted $P(X_i + Y_i \leq 1)$ for $0 \leq \lambda \leq 5$. Does this have a simple form?

enter image description here

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  • $\begingroup$ How did $C_Z$ look like before you have fitted it? $\endgroup$ – callculus Sep 25 at 16:43
  • $\begingroup$ @callculus I have plotted $C_Z$ as the histogram. The only fitting is the normal distribution line I have drawn just to show that it seems to be approximately normal (and hence approximately binomial). $\endgroup$ – Anush Sep 25 at 16:47
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I don't know what you mean by shifted, but assuming that $Y_i$s all have the same parameter for their distribution, the random you are looking for will be binomial with parameters (p, c) with $p = P(X_i + Y_i \leq 1)$. This is because you can think of each event $\{X_i + Y_i \leq 1 \}$ as an independent trial (say a coin toss), and you will do it $c$ times.

Note: In the question it currently says $X_i + Y$ instead of $X_i + Y_i$ but from context it seems the intention was to ask $X_i + Y_i$ so I answered it for $X_i + Y_i$; if the intended meaning is $X_i + Y$, then it will not necessarily be binomial; its distribution would be the average of binomial distributions.

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  • $\begingroup$ Thank you. What is $P(X_i + Y_i \leq 1)$? $\endgroup$ – Anush Sep 25 at 18:21
  • $\begingroup$ Is it $(-1 + e^{-\lambda} + \lambda)/\lambda$? $\endgroup$ – Anush Sep 25 at 19:30

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