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I stumbled against this problem and found it really hard and help would be much appreciated

Let $X_{1},X_{2},......,X_{n}$ be a series of independent Bernoulli variables with $P(X_{i}=1)=\theta$ and $P(X_{i}=0)=1-\theta$

Let $Y_{n}= X_{1}+X_{2}+......+X_{n}$

I know that $Y_{n}$ is a Binomial distribution with $n \theta$ as expected value and $n \theta (1-\theta)$ as variance but I can't figure out the probability distribution of $n-Y_{n}$ and how to compute its expected value and variance

A second question supposed that $n \sim \mathcal{Poisson} (\lambda)$ compute the expected value and variance of $Y_{n}$

The puzzling thing for me how $n$ as an index can change the distribution

Thank you for your time

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  • $\begingroup$ You could assume for a while, that $\theta$ is equal to 1/2, and experiment with a coin. That values lets say $Y_3$ or $Y_4$ may take? What values happen more often? Also, some more clarity in the question would be appreciated. $\endgroup$ – cure Oct 18 at 11:01
  • $\begingroup$ Made some mistakes and I fixed them sorry if the problem is not clear English is not my first language I don't know how to find the probability distribution of $n-Y_{n}$ The second part is if $n$ is a Poisson distribution what is the expected value and variance of $Y_{n}$ $\endgroup$ – chaabouni ali Oct 18 at 11:14
  • $\begingroup$ You ask 'second question', however I have some troubles with finding and understanding the first one. Also, please have in mind, that self-study questions are different than others: stats.stackexchange.com/tags/self-study/info $\endgroup$ – cure Oct 18 at 12:13
  • $\begingroup$ Sorry I will try to explain as much as I could We have $n$ independent Bernoulli variables Their sum is $Y_{n}$ which is a Binomial distribution What is the probability distribution of $n-Y_{n}$ $\endgroup$ – chaabouni ali Oct 18 at 13:45
  • $\begingroup$ For the first question try to assume some value of $n$ and $\theta$. For example 3 and 1/2. You can try to calculate probability, that in 3 coin tosses there will be 0, 1, 2, 3 heads. Then, for every situation, what are probabilities of 3 - the number of heads? How will it change when $n$ is different? Can you generalise it? $\endgroup$ – cure Oct 18 at 14:05
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For you second question, since you already know that the mean of $Y_{n}$ is $n\theta$, and $n\sim\mathcal{Poisson}(\lambda)$, the mean of $Y_{n}$ is

$E[Y_{n}] = E[n\theta]=\theta E[n]=\lambda\theta$. Or equivalently, $E[Y_{n}]=\sum_{n} n\theta*\frac{\lambda^{n} e^{-\lambda}}{n!}=\lambda\theta$.

Similarly, the various of $Y_{n}$ is $\sum_{n} n\theta (1-\theta)*\frac{\lambda^{n} e^{-\lambda}}{n!}=\lambda\theta (1-\theta)$.

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For the first question, since $Y_{n}$ is of binomial distribution, the probability that $Y_{n}=y$ is

$P(Y_{n}=y)={n \choose y}\theta ^y\times(1-\theta)^{n-y}$.

The probability that $n-Y_{n}=k$ then is

$P(n-Y_{n}=k)=P(Y_{n}=n-k)={n \choose n-k}\theta ^{n-k}\times(1-\theta)^{k}={n \choose k}(1-\theta) ^{k}\times\theta^{n-k}$.

Thus, $n-Y_{n}$ is of binomial distribution with parameters of $(n;1-\theta)$. In other words, while $Y_{n}$ is the number of $Xi=1$, $n-Y_{n}$ is the number of $Xi=0$, among $n$ tests.

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