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Suppose that $X_1,\cdots,X_n$ are jointly continuous with joint pdf $$f(x_1,\cdots,x_n) = (2\pi)^{-n/2} \exp\left\{-\frac{1}{2} \left[x_n^2+\sum_{i=1}^{n-1}(x_i-x_n)^2\right]\right\}.$$ Find the joint marginal pdf of $(X_1,\cdots,X_n)$. Are $X_1,\cdots,X_n$ mutually independent?

I know that to find the marginal, I'm going to need to integrate the pdf above over all $x_n$, i.e. over $\mathbb{R}$. But this integral looks pretty messy to me. The full joint pdf looks a lot like some kind of normal distribution. I was hoping to somehow use that fact so that part of what I'm integrating would end up just being $1$.

Any thoughts on how to approach this? Thanks!

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  • $\begingroup$ At one point you seem to call the number of random variables $N$ and in another you call it $n$. Are those both supposed to be the same thing? If so, you should choose one of the two and stick with it. In standard usage, $n$ and $N$ are two different things. ${}\qquad{}$ $\endgroup$
    – Michael Hardy
    Commented Jan 14, 2015 at 17:52
  • $\begingroup$ @MichaelHardy Ah, yes, that $N$ is supposed to be $n$. Just a matter of a lazy finger on the Shift key after typing the underscore. ;) $\endgroup$
    – kokocijo
    Commented Jan 14, 2015 at 20:04

1 Answer 1

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Hint 1: Condition on $X_n$ first to see what the structure is.

Hint 2: $X_{1} , \ldots, X_{n-1}$ are i.i.d. $N(X_n, 1)$ given $X_n$, which is $N(0,1)$.

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  • $\begingroup$ So, writing it as $$f(x_1,\dotsc,x_n) = (2\pi)^{-1/2}\:\exp{\left(-\frac{x_n^2}{2}\right)} \: \prod_{i=1}^{n-1}(2\pi)^{-1/2}\:\exp{\left(-\frac{(x_i-x_n)^2} {2}\right)}$$ shows that we can set $$f_{X_n}(x_n):=(2\pi)^{-1/2}\:\exp{\left(-\frac{x_n^2}{2}\right)}$$ and the rest similarly, just by recognizing the form of the pdf of the Gaussian distribution? $\endgroup$
    – kokocijo
    Commented Jan 15, 2015 at 2:25
  • $\begingroup$ Yes. You're just recognizing $f(x_1,\ldots,x_n) = f(x_1,\ldots,x_{n-1} |x_n) f(x_n)$ then recognizing $f(x_1,\ldots,x_{n-1}|x_n) = f(x_1|x_n)\ldots f(x_{n-1}|x_n)$ by factoring. $\endgroup$
    – Batman
    Commented Jan 15, 2015 at 12:43
  • $\begingroup$ And f(x_n) is the $N(0,1)$ pdf and $f(x_i | x_n)$ is the $N(x_n, 1)$ pdf for $i<n$. $\endgroup$
    – Batman
    Commented Jan 15, 2015 at 13:11

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