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Let $Y_1,..,Y_n$ be random variables. Given that for all $1\le k \le n-1,k\in \mathbb{N}$, $Y_k$ is independent to the joint distribution of the other $n-1$ random variable, prove that $Y_n$ is also independent of the other $n-1$ random variables. I have tried to show that all $Y_1,...,Y_{n-1}$ are independent but not sure how to show that even $Y_n$ are independent of joint distributions $Y_1,...,Y_{n-1}$.

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What happened to this question? –  Did Oct 29 '12 at 11:09

2 Answers 2

Let $\phi_k$ be bounded functions with the support on the domain of the corresponding $Y_k$. Consider: $$ f(Y_1, Y_2, \ldots, Y_{n-1}, Y_n) = \phi_1(Y_1)\phi_2(Y_2) \cdots \phi_{n-1}(Y_{n-1}) \phi_n(Y_n) $$ The independence of $Y_1$ from the remaining random variables means that: $$ \mathbb{E}\left(\phi_1(Y_1)\phi_2(Y_2) \cdots \phi_{n-1}(Y_{n-1}) \phi_n(Y_n)\right) = \mathbb{E}\left(\phi_1(Y_1)\right) \cdot \mathbb{E}\left(\phi_2(Y_2) \cdots \phi_{n-1}(Y_{n-1}) \phi_n(Y_n)\right) $$ Since $Y_2$ is also independent of the other $n-1$, we get: $$ \mathbb{E}\left(\phi_1(Y_1)\phi_2(Y_2) \cdots \phi_{n-1}(Y_{n-1}) \phi_n(Y_n)\right) = \mathbb{E}\left(\phi_1(Y_1)\right) \cdot \mathbb{E}\left(\phi_2(Y_2)\right) \cdot \mathbb{E}\left(\phi_3(Y_3) \cdots \phi_{n-1}(Y_{n-1}) \phi_n(Y_n)\right) $$ Continuing for $k \leqslant n-1$ we get: $$ \mathbb{E}\left(\phi_1(Y_1)\phi_2(Y_2) \cdots \phi_{n-1}(Y_{n-1}) \phi_n(Y_n)\right) =\prod_{k=1}^{n-1} \mathbb{E}\left(\phi_{k}(Y_k)\right) \cdot \mathbb{E}\left(\phi_n(Y_n)\right) = \prod_{k=1}^{n} \mathbb{E}\left(\phi_{k}(Y_k)\right) \tag{1} $$ Since eq. $(1)$ holds for arbitrary bounded functions $\phi_k$, it implies that $\{Y_k\}$ are independent of one another and, in particular, that $Y_n$ is independent of the previous $n-1$ random variables $Y_1,\ldots, Y_{n-1}$.

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Assume $n=3$. Then $Y_2$ is independent of $(Y_1,Y_3)$, in particular $Y_2$ is independent of $Y_3$. And $Y_1$ is independent of $(Y_2,Y_3)$, hence $Y_1$, $Y_2$ and $Y_3$ are independent. The case $n\geqslant4$ is similar.

Here is a way to really learn something from the question: justify each step of the (indications of) proof given above for the case $n=3$. If a step stays unclear, just say so.

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then use mathematical induction to prove n right?? –  Mathematics Oct 8 '12 at 15:50
?? Provided you have a clear idea of the recursion you would be using. –  Did Oct 8 '12 at 15:53
but we only know $Y_2$ is indepedent to the joint pmf of $Y_1,Y_3$ instead of $Y_1,Y_3$ –  Mathematics Oct 8 '12 at 16:07
A random variable is not independent to the joint probability distribution of other random variables, a random variable is (or is not) independent of other random variables (defined on the same probability space). So... what do you think your hypothesis means exactly? –  Did Oct 8 '12 at 16:11
i guess it is saying may be the pmf of $Y_2$ and joint pmf of $Y_1,Y_3$ –  Mathematics Oct 8 '12 at 16:13

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