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Claim. Let $Z_1, Z_2$ be two independent and identically distributed random variables. Then we have: $$ \mathbb E[Z_1|Z_1+Z_2] =\frac{Z_1+Z_2}{2}. $$ Proof. To see this, I have proceeded as follows. From the general properties of conditional expectation, we have: $$ \mathbb E[Z_1+Z_2|Z_1+Z_2] =Z_1+Z_2. $$

Now, again from the general properties of c.e. (linearity), I can write: $$ \mathbb E[Z_1+Z_2|Z_1+Z_2] =\mathbb E[Z_1|Z_1+Z_2] +\mathbb E[Z_2|Z_1+Z_2] =2\mathbb E[Z_1|Z_1+Z_2] $$ The last equality because $Z_1$ has the same distribution as $Z_2$. Putting the two equalities together we find immediately the identity stated in the Claim.

Question What I don't find clear is where exactly the condition of independence of the two random variables is used. I know that the result is not true if they aren't independent, but I don't see where this conditoin is needed. The only point I can think of is in the last step of the chain of equalities, where we use the fact that $\mathbb E[Z_1|Z_1+Z_2]=\mathbb E[Z_2|Z_1+Z_2]$ but it seems to me that this holds just beacuse of the same distribution of $Z_1, Z_2$ and does not require independence.

And yet, the result is not true if $Z_1, Z_2$ are not independent. So, where is the independece condition used?

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    $\begingroup$ You should mention in the first line that $Z_1$ and $Z_2$ are identically distributed. $\endgroup$ – joriki Jun 28 '16 at 11:25
  • $\begingroup$ Also note that you get proper spacing by using \mid instead of `|'. $\endgroup$ – joriki Jun 28 '16 at 11:26
  • $\begingroup$ Yes, thanks! :) $\endgroup$ – RandomGuy Jun 28 '16 at 11:28
  • $\begingroup$ To repeat what others already said: if $(X,Y)$ is distributed as $(U,V)$ and if $E(X\mid Y)=g(Y)$ then $E(U\mid V)=g(V)$. In your case, $(Z_1,Z_2)$ is i.i.d. hence $(Z_1,Z_2)$ is exchangeable (that is, $(Z_1,Z_2)$ and $(Z_2,Z_1)$ have the same distribution) hence $(Z_1,Z_1+Z_2)$ and $(Z_2,Z_1+Z_2)$ have the same distribution, qed. $\endgroup$ – Did Jun 28 '16 at 14:32
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We need only show that, for any Borel set $A \in \mathbb{R}$, \begin{align*} \int_{Z_1+Z_2 \in A} Z_1 dP = \int_{Z_1+Z_2 \in A} Z_2 dP. \end{align*} We denote by $F$ the common cumulative distribution function of $Z_1$ and $Z_2$. Then, from the independence assumption, \begin{align*} \int_{Z_1+Z_2 \in A} Z_1 dP &= E\left(Z_1 \pmb{1}_{Z_1+Z_2 \in A} \right)\\ &=\iint_{\mathbb{R}^2} x \pmb{1}_{x+y \in A} dF(x) dF(y). \end{align*} Analogously, \begin{align*} \int_{Z_1+Z_2 \in A} Z_2 dP &= E\left(Z_2 \pmb{1}_{Z_1+Z_2 \in A} \right)\\ &=\iint_{\mathbb{R}^2} u \pmb{1}_{v+u \in A} dF(u) dF(v)\\ &=\iint_{\mathbb{R}^2} u \pmb{1}_{u+v \in A} dF(u) dF(v). \end{align*} That is, \begin{align*} \int_{Z_1+Z_2 \in A} Z_1 dP = \int_{Z_1+Z_2 \in A} Z_2 dP. \end{align*} In other words, \begin{align*} E\left(Z_1 \mid Z_1+Z_2 \right) = E\left(Z_2 \mid Z_1+Z_2 \right). \end{align*}

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  • $\begingroup$ I have a question. I understand your neat proof, however, there is a point I would like to clarify. If $Z_1, Z_2$ were not independent, you could still write: $E(Z_1 \pmb{1}_{Z_1+Z_2 \in A} )=\iint_{\mathbb{R}^2} x \pmb{1}_{x+y \in A} dF(x,y)$, and similarly the other integral. But why this would not be enough to prove that the two integrals are equal? And further need to factor out the measure as $dF(u)dF(v)$ to prove it? $\endgroup$ – RandomGuy Jun 29 '16 at 10:16
  • $\begingroup$ That is true,but you are not able to switch the order of x and y. $\endgroup$ – Gordon Jun 29 '16 at 10:19
  • $\begingroup$ Only if you are able to switch the order of x and y, or u and v, you can say that the two integrals are the same $\endgroup$ – Gordon Jun 29 '16 at 10:26
  • $\begingroup$ in other words, if I understand well, you are saying that the two integrals $\iint_{\mathbb{R}^2} x \pmb{1}_{x+y \in A} dF(x,y)$ and $\iint_{\mathbb{R}^2} y \pmb{1}_{x+y \in A} dF(x,y)$ are in general different, is it so? But if you add the independence condition, then by writing the two as iterated integrals, they become equal as you wrote, is it correct? $\endgroup$ – RandomGuy Jun 29 '16 at 10:43
  • $\begingroup$ In the two integrands, the positions of x and y are different. If you add independence, you can then switch the order and two integrands become the same. $\endgroup$ – Gordon Jun 29 '16 at 11:02
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In addition to the identical distribution independence is used to obtain that $(Z_1,Z_1+Z_2)$ and $(Z_2,Z_1+Z_2)$ are identically distributed which is needed to conclude $\mathbb E(Z_1 \mid Z_1+Z_2) = \mathbb E(Z_2 \mid Z_1+Z_2)$.

Edit. The distribution of $(Z_1,Z_1+Z_2)$ is the image (push forward) of $\mathbb P^{(Z_1,Z_2)}= \mathbb P^{Z_1} \otimes \mathbb P^{Z_2}$ (here independence is used) under the transformation $(u,v)\mapsto (u+v,v)$.

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  • $\begingroup$ Can you explain why one needs independence to make sure that $(Z_1, Z_1+Z_2)$ and $(Z_2, Z_1+Z_2)$ are identically distributed? $\endgroup$ – RandomGuy Jun 28 '16 at 11:47
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    $\begingroup$ @RandomGuy: The example in my answer shows that one does. It would be your turn now to explain why you think that one doesn't. In the question, you had several steps, and we pointed out the incorrect step. This time you're not providing any steps by which you suggest to prove that this holds without independence, so there's nothing for us to point out. You're merely claiming something wrong without providing an argument for it that we could disprove. $\endgroup$ – joriki Jun 28 '16 at 13:08
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Let $(Z_1,Z_2)$ be uniformly distributed on $\{(-1,0),(0,1),(1,-1)\}$. Then $Z_1$ and $Z_2$ are identically distributed but dependent. The value of $Z_1+Z_2$ fixes the (different) values of $Z_1$ and $Z_2$, so $\mathbb E[Z_1\mid Z_1+Z_2]\ne\mathbb E[Z_2\mid Z_1+Z_2]$.

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  • $\begingroup$ Ok. But my question is where exacty in the proof the independence condition is used. $\endgroup$ – RandomGuy Jun 28 '16 at 12:48
  • $\begingroup$ @RandomGuy: But that's what I showed. Since $\mathbb E[Z_1\mid Z_1+Z_2]\ne\mathbb E[Z_2\mid Z_1+Z_2]$ for dependent variables, you used the independence condition when you assumed $\mathbb E[Z_1\mid Z_1+Z_2]=\mathbb E[Z_2\mid Z_1+Z_2]$. It's the same answer as Jochen's, just with an example. $\endgroup$ – joriki Jun 28 '16 at 13:03
  • $\begingroup$ yes, I see your point. Sorry, I had completely misunderstood the meaning of your post. Thank you $\endgroup$ – RandomGuy Jun 29 '16 at 7:56
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Just as another remark: a neat corollary which follows from this statement and the strong law of large numbers is that whenever $X_i$ is an iid sequence of r.v. such that the SLLN holds (e.g. $X_i \in L^1$ by Kolmogorov's SLLN), then $\mathbb{E}(X_1 \mid S_n) \to \mathbb{E}(X_1)$ $\mathbb{P}$-almost surely.

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