Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
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? –  Srivatsan Jul 28 '11 at 18:53
    
Are you referring to the first or second question? –  Godwin Jul 28 '11 at 19:00
    
Actually, both. I am hoping the word "probability distribution" is used in a consistent sense in this question. –  Srivatsan Jul 28 '11 at 19:04
    
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 $x_j$. –  André Nicolas Jul 28 '11 at 19:27
1  
@Srivatsan: There's no such thing as a "cumulative density function". The words "cumulative" and "density" contradict each other. –  Michael Hardy Jul 28 '11 at 21:54

1 Answer 1

We will assume that the $X_j$ are independent. This assumption is not automatically built into the definition of random sampling, but it is necessary if we are to give a complete answer.

If $F(x)$ is the cumulative distribution function for the population, and $F_n(x_1, x_2,\dots, x_n)$ is the joint sample (cumulative) distribution function, then, more or less as you wrote, we have $$F_n(x_1,x_2,\dots,x_n)=P(X_1 \le x_1)(P(X_2\le x_2)\cdots P(X_n \le x_n).$$ Please note that $F_n$ is a function of the $n$ real variables $x_1, x_2, \dots,x_n$. (No caps!) The reasoning that you used to get to this stage was correct.

We can therefore write $$F_n(x_1,x_2,\dots,x_n)=F(x_1)F(x_2)\cdots F(x_n).$$

The final displayed expression in the post, namely $[P(X_1 \le x_1)]^n$, is not correct, and cannot be correct, for it does not mention the variables $x_2$ to $x_n$. In the form $[F(x)]^n$, it does occur in the calculation of the distribution of the largest sample value, but that is not the problem you were asked to look at.

If we want the joint density function $f(x_1,x_2,\dots,x_n)$, we just multiply the individual density functions $f(x_j)$.

The (marginal) distribution of any $X_j$ is, by independence, the same as the population distribution. So if you want to specify the distribution by using a cdf, the answer would be simply $F(x_j)$. If we are in a continuous situation, and $F'(x)=f(x)$, then the (marginal) density function of $X_j$ is $f(x_j)$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.