3
$\begingroup$

I have a r.v $S_N$ built as a sum of Bernoulli with parameter $p$. So $S_N = X_1 + X_2 + \ldots + X_N$. There is a second variable N, such that $N \sim Poisson(\lambda) $.

I have to compute:

  1. $P(S_N=0)$
  2. $\mathop{\mathbb{E}}(S_N \ | \ N = 4 )$
  3. $\mathop{\mathbb{E}}(S_N \ | \ N )$

Now, for the first point, I need to think my sum of Bernoulli as a Binomial. Thus:

$$ P(S_N = 0) = \binom{n}{0} p^k (1-p)^{n-k} = (1-p)^n $$

However, i'm stuck with the two expectations. But i remember that for the conditional expectaction for two discrete random variables it holds:

$$ \mathop{\mathbb{E}}(X \ | \ Y = k ) = \sum_x x \ f_{X|Y}(x|y) $$

$\endgroup$
9
  • 2
    $\begingroup$ I am confused. Are $Y$ and $N$ the same? $\endgroup$ – Siong Thye Goh Nov 12 '20 at 17:15
  • 1
    $\begingroup$ What is the role of the second variable $Y$ should it be $N$? $\endgroup$ – Sextus Empiricus Nov 12 '20 at 17:17
  • 3
    $\begingroup$ Since $N$ is random, $S_N$ cannot be a Binomial $\mathcal B(N,p)$. $\endgroup$ – Xi'an Nov 12 '20 at 17:27
  • 2
    $\begingroup$ See stats.stackexchange.com/search?q=poisson+binomial+expectation. $\endgroup$ – whuber Nov 12 '20 at 17:47
  • 1
    $\begingroup$ $S_N$ will be Poisson with rate $\lambda p$. With this you can answer question 1 (but probably you are supposed to compute it manually). $\endgroup$ – Sextus Empiricus Nov 12 '20 at 19:18
1
$\begingroup$

In Problem 2, $N=4$ is a constant. Thus, $S_N$ is a binomial variable with the mean equals $4p$ (assuming the Bernoulli takes value either 1 or 0).

For Problem 3: In general for the binomial distribution with $N=n$, the mean is $np$. Now for $N$ is a random variable, the mean then is $E[np]=\bar{n}p$. In particular when $N$ is of Poisson distribution, $\bar{n}=\lambda$, thus the mean is $\lambda p$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.