I am a bit confused on the concept of conditioning and finding the new distribution of conditioned random variables.

An example,

Suppose that there is a new carnival in town and that citizens will begin to line up outside at $10:00$ am. Further suppose that the arrivals are independent, and that the number that arrive in any 1 minute interval is Poisson distributed with $\lambda=3$.

Now suppose the owners will check on the line $T$ minutes after people begin to show up, where $T$ is binomial distributed with $n=20$ and $p=0.8$, then find the expected number of citizens in the line when it is checked and the standard deviation.

What I tried:

I know the law of double expectation and variance

the first is that $E[P]=E[E(P\mid T)]$

and $Var(P)=E[Var(P\mid T)]+Var[E(P\mid T)]$ which I could find standard deviation from.

I know that $E(T)=np=16$ and $E(P)=\lambda=3 $

If I had to give an answer without being confident, I would say that the expected number of customers would be 48

So I know that $E(C|T)$ ( I now think it is $3T$) should be a random variable, how do I find its distribution? What about in general to find the distribution of conditional expectation?

I am having some trouble on the second part and this is why, to use the formula I listed for variance above I want to calculated $Var(P|T)$ , I thought this would be equal to $E(P^2|T)-(E(P|T))^{2}$ which would give me $9T-9T^{2}$, which has an expected value of $-2188.8$, adding this to the variance of the expected value $( 28.8)$ still leaves a very negative number which I know cannot be the case. so I must have a mistake somewhere. Here I also used that we know the variance of binomial to be npq , which in this case is =3.2 and used this to solve for $E(M^2)$ I have tried over and over now but cant figure out where and why I am not getting an answer that makes sense.

Still confused, is really no one willing to help clarify this at all?? I am also confused in one of the answers how he says that the r.v is poisson



You should have $\operatorname{E}(P\mid T) = \lambda T.$

Remember that $\lambda$ is in persons-per-minute and $T$ is in minutes, so when you multiply them you get persons.

So $\operatorname{E}(\operatorname{E}(P\mid T)) = \operatorname{E}(\lambda T) = \lambda\operatorname{E}(T) = \cdots$ etc.

And $\operatorname{var}(P\mid T) = \lambda T$ (since the variance of a Poisson distribution is the same as the expected value.

  • $\begingroup$ Thanks, can we say $Var(P|T)=E(P^2|T)-(E(P|T))^2$ and wouldn't $E(P^2|T)=9T$ while $E(P|T)=3T$ as we saw so variance would be $9T-9T^{2}$, where is the mistake in thinking that? $\endgroup$ – PersonaA Jan 10 '17 at 20:55
  • $\begingroup$ Also I get that variance is same for Poisson, but is $\lambda$ T poission? I know $\lambda=3$ and $T$ is binomial, I thought $3T$ would be binomial as well $\endgroup$ – PersonaA Jan 10 '17 at 20:57
  • $\begingroup$ Why is P|T poisson? $\endgroup$ – PersonaA Jan 11 '17 at 19:45
  • $\begingroup$ @PersonaA : Conventionally when one speaks of a Poisosn distribution in this way, the numbers of people arriving in different time intervals that don't overlap are independent. Thus $P$ is the sum of $T$ independent Poisson-distributed random variables, each with expected value $3$. So $P \mid T\sim\operatorname{Poisson}(3T). \qquad$ $\endgroup$ – Michael Hardy Jan 11 '17 at 20:06
  • $\begingroup$ @PersonaA : Neither $\lambda T$ nor $3T$ has a Poisson distribution, nor does either $\lambda T$ or $3T$ have a binomial distribution. But the conditional distribution of $P$ given $T$ is Poisson with expected value $3T$. Since the variance of a Poisson distribution is the same as its expected value, the variance of $P$ given $T$ is also $3T. \qquad$ $\endgroup$ – Michael Hardy Jan 11 '17 at 20:09

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