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Suppose $X$ is a random point in the Euclidean space $R^3$ with some non-discrete distribution, for example, $X$ is uniform in a bounded area. $U$ is another random point which is uniform on the sphere $B(0,1)$ and independent of $X$, set $Y=X+U$. Set $Z_1=1/|X|$ and $Z_2=1/|Y|$, where $|X|$ is the distance of $X$ to the origin, $|Y|$ similarly. We want to compute the conditional expectation $E[Z_2|Z_1]$.

The computation goes as follows: If we know $Z_1=a$, then $Y$ is a uniform random point on the sphere S, where S has radius 1, and the distance between the center of S and the origin is $1/a$. Use some calculus, one can compute the value of $Z_2$ given the information $Z_1=a$:

$$E[Z_2|\ \{|Z_1|=a\}]=a\quad \text{if} \ a<1$$

and

$$E[Z_2|\ \{|Z_1|=a\}]=1\quad\text{if}\ a\geq 1$$

So it concludes that $E[Z_n|Z_{n-1}]\leq Z_{n-1}$. (supermartingale).

I can not understand why one can compute the conditional expectation with respect to a continuous random variable as above: If we want to compute $E[Z|Y]$ when $Y$ is a continuous r.v. and for every possible value $Y=y$, the conditional expectation of $Z$ given the event $E[Z|\{Y=y\}]$ happens to exist and can be computed (denote as $f(y)$), then one can assert that$$E[Z|Y]=f(Y).$$

I know this way coincides with the intuition of conditional expectation; but I can't see why it agrees with the formal definition of conditional expectation: for every event $A$ in $\sigma(Y)$, $$\int_A f(Y) \, dP=\int_A Z \, dP.$$

Can anyone explain why? Thanks.

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    I dont see how you can have: $$E[Z_2|\ \{|Z_1|=a\}]=a\quad \text{if} \ a<1$$ and $$E[Z_2|\ \{|Z_1|=a\}]=1\quad\text{if}\ a\geq 1$$ Let's say $X$ is uniform in $B(O,r)$, with $r<1$, you will then have $Z_1>1$, but I dont see why it implies that $E(Z_2|Z_1) = 1$.2012-08-13
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    You might want to explain what you call $E[Z|\{Y=y\}]$ when $\{Y=y\}$ has probability zero and how you compute this number without resorting to the random variable $E[Z|Y]$. This precise point is probably at the heart of the question.2012-08-13
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    @did: this is an intuitive way to derive the conditional expectation: given the position of $Z_1$, one computes the expectaion value of $Z_2$ in this situation (a function of $Z_1$) and take this as the real conditional expectation. This is just why I can not understand.2012-08-13
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    Please tell me if my answer below helps.2012-08-13
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    @S4M These values are correct (and follow from the fact that $x\mapsto1/|x|$ is harmonic on $\mathbb R^3\setminus\{0\}$).2012-08-14

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