Urgent :problem in deriving mean for Compound distribution

[Arpan]

E= E[E[S|N]]

E[S|N]= E[X(1) + X(2) +....+ X(N) | N = n ]

E[S|N]= E[X(1) + X(2) +....+ X(n) ] --(1) (Rest of the steps fit logically ultimately getting E = E[X]. E[N])

where does condition get vanished in (1)? To my mind, in this we have conditioned on the particular value of N i.e. n therefore the condition has gone out as it's no longer a random variable.

If not this then kindly give more plausible explanation.

 

[DevonMatthews]

The condition doesn't vanish, you just chose some arbitrary realisation of the random variable N to work with, in this case "n".

So E[S|N=n]= E[X(1) + X(2) +....+ X(n) ] = nE[X] since X1, X2 ... Xn are independent and identically distributed.

But rather than chose an actual number, just condition on knowledge of the random variable by writing an analogous expression to the one above and then average over all it's values. ie. E[S|N] = NE[X]. Now using the unconditional expectation result E = E[E[S|N]] = E[NE[X]] = E[N]E[X].

 

[Simon C]

Confused by N

I'm resurrecting this old thread on compound distributions as it touches upon something I'm stuck on.

1)
I get that E[S | N = n] = E[X(1) + X(2) + ... + X(n) | N = n]
= E[X(1) + X(2) + ... + X(n)]
= nE[X]

Using summation formulae, I can also prove that E(E(S | N)) = E(N)E(X).
However, the Core Reading does the proof in a different way where we have E(E(S | N)) = E(NE(X)) = E(N)E(X). How do we get the N by itself inside the expectation of the second of these expressions?

DevonMatthews has touched on this in the penultimate sentence of their post below but I'm not quite getting it. Would be grateful if someone could clarify.

I'm similarly confused by the N in the second of the expressions in E(var(S|N)) = E(Nvar(X)) = E(N)var(X).

2) How is var(E(S | N)) = var(NE(X | N))?

Thanks
Simon

 

[freddie]

The Core Reading is using the connection between unconditional expectation and conditional expectation. If we want to find the unconditional expectation of S, ie E, we can take the expectation of the conditional expectation of S, ie E[E[S/N]].

As you say, you know the conditional expectation of S, ie E[S/N] = N E[X], so to find the unconditional expectation of S we take the expectation of the conditional expectation, ie E = E[E[S/N]] = E [N E[X]]. N is now the random variable and E[X] is a constant, so E = E[N] E[X].

In the same way, the conditional variance of S, ie var[S/N] = N var[X], so, using the formula on page 16 of the Tables we can find the unconditional variance of S.

var = E[var[S/N]] + var [E[S/N]] = E[N var[X]] + var [N E[X]]. Then usng expectation and variance rules, and treating N as the random variable, we arrive at E[N] var[X] + [E[X]]^2 var[N].

 

[Simon C]

Thanks very much for the reply Freddie. Possibly I haven't explained where I'm getting confused very well as I'm still confused on the same point

I understand and can prove that E(E(S | N)) = E(N)E(X) using sigma notation.

But in the Core Reading's method for proving the same thing, I don't follow how we move from E(E(S | N)) to E(N(E(X)).

I follow that E[S | N = n] = nE[X]. However I don't get what we are saying when we say E(S | N) = NE(X). Would be grateful for any clarification you or anyone else can provide.

Thanks again
Simon

 

[freddie]

I'm sorry I missed your point! Actually I think DevonMatthews has already answered it. You can condition the expectation on a particular value of N, ie E[S/N=n] = nE[X] (in which case we would have a particular value) or you can condition the expectation for any value of the random variable N, ie E[S/N] = NE[X] (in which case we have a function of the random variable N). I hope this helps!

 

[Simon C]

Thanks Freddie, I think I am just about getting it now.

Thinking of NE(X) as a function of random variable N is what has helped