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Let $X_{1},X_{2},\ldots ,X_{n}$ be a random sample of size $n$ from a population distribution $F$. I want to find the following:
1. the joint P.d.f of $X_{1},X_{2},\ldots ,X_{n}$.
2. the marginal probability distribution of $X_{j}$ for any $j$ in $1\leq j \leq n$.


This is my attempt for 1.
Given $F$, the joint pdf of the random sample is
$$\begin{align*} F\left(X_{1},X_{2},\ldots ,X_{n}\right) & =P \left(X_{1}\leq x_1, X_{2}\leq x_2, \ldots X_n\leq x_n \right)\\ &=P(X_1 \leq x_1)P(X_2\leq x_2)\dot{} \ldots \dot{}P(X_n\leq x_n) \\ &=\left[P(X_1\leq x_1)\right ]^n \qquad \because X_j\text{'s are identical} \end{align*} $$


Here are my questions: First, iIs my attempt for 1. right. Is there a better way of doing it.Second, I would like some help with 2. I know the $X_j$'s will all have the same marginal distributions, but I don't know how to justify it.

Thanks.

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    This PDF looks like the **cumulative** density function (CDF) for the random variable $\max \{ X_1, X_2, \ldots, X_n \}$, rather than the joint PDF. BTW by probability distribution, do you mean PDF or CDF?2011-07-28
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    Are you referring to the first or second question?2011-07-28
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    Actually, both. I am hoping the word "probability distribution" is used in a consistent sense in this question.2011-07-28
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    The last line (the one with "$\dots$ identical") is completely wrong. If we have independent random sampling, say from a continuous distribution, the joint density is the product of the individual densities, which in this context are called marginal densities. Keep the different variables $x_j$.2011-07-28
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    If $F(x)$ is the population cdf, and $f(x)$ is the population pdf, then the marginal pdf of $X_j$ is $f(x_j)$, by independence.2011-07-28
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    @Srivatsan: There's no such thing as a "cumulative density function". The words "cumulative" and "density" contradict each other.2011-07-28
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    @Michael Yes, "cumulative density function" is wrong, thanks for pointing it out. (The funny thing is I pretty much always use CDF, but here I used the expanded form since I wanted to emphasize the word "cumulative". :)) But my doubt/confusion still remains, with "cumulative density function" replaced by "cumulative distribution function".2011-07-29

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