Random Variable Transformation

[SupLem]

We have a r.v. X with p.d.f. = sqrt(θ/πx)*exp(-xθ) , x>0 and θ a positive parameter.
We are required to show that 2 θX has a x^2 distribution with 1 d.f. and deduce that, if x_1,……,x_n are independent r.v. with this p.d.f., then 2θ∑_(i=1)^n▒x_ι has a chi-squared distribution with n degrees of freedom.
Using transformation (y=2θΧ) I found the pdf of y=1/sqrt (2π) *y^(-1/2)*e^(-y/2). How do I find the distribution of 2θ∑_(i=1)^n▒x_ι ? Do I need to find the likelihood function (which contains ∑_(i=1)^n▒x_ι ) first ? How do I recognise the d.f. of this distribution (Is it n because it involves x_1,……,x_n,i.e. n r.v.?

(since i couldn't get the graphics right above, I am also adding a screenshot of my word document in order to view). Thanks!View attachment 3491

 

[Siron]

As you've found for the pdf $f_Y$ of the r.v $Y=2\theta X$:
$$f_Y(y) = \frac{1}{\sqrt{2\pi}} y^{-\frac{1}{2}} e^{-\frac{y}{2}}$$
which is indeed the pdf of the $\chi^2$-distribution with 1 degree of freedom. Next, we have given that $X_1,\ldots,X_n$ are independen r.v's with the same pdf as $X$.

First note that $2\theta X_i$ for $i=1,\ldots,n$ has the same distribution as $Y$, in other words to solve the question you have to find the distribution of $n$ independent r.v's $X_i$ with $X_i \sim \chi^2(1)$.

Do you need to solve this with the transformation theorem? Because using the moment-generating function would lead easily to a solution (because of the independency).

 

[SupLem]

Thank you very much for your response. Could you, please, elaborate, on how using the moment generating function would help us in this respect ( i.e. finding the 2θΣx(i=1 to n) distribution?

 

[Siron]

Thank you very much for your response. Could you, please, elaborate, on how using the moment generating function would help us in this respect ( i.e. finding the 2θΣx(i=1 to n) distribution?

It satisfies to use the moment generating function as it determines the distribution completely.

Denote the moment generating function of $2\theta X_i \sim \chi^2(1)$ as $M_{2\theta X_i}(t)$ which is known for $i=1,\ldots,n$. Due to the independency we have
$$M_ {2\theta \sum_{i=1}^{n} X_i}(t) = \prod_{i=1}^{n} M_{2\theta X_i}(t)$$

 

[blue_raver22]

I would like to provide a response to the content regarding random variable transformation and the derivation of the distribution of 2θ∑_(i=1)^n▒x_ι.

Firstly, we are given a random variable X with a probability density function (p.d.f.) of sqrt(θ/πx)*exp(-xθ), where x>0 and θ is a positive parameter. We are required to show that 2θX has a chi-squared distribution with 1 degree of freedom. In order to do this, we can use the transformation method.

We define a new random variable, Y=2θX, and substitute it into the given p.d.f. for X. This gives us the p.d.f. for Y as 1/sqrt(2π)*y^(-1/2)*exp(-y/2). This is the p.d.f. of a chi-squared distribution with 1 degree of freedom, as can be seen by comparing it to the general form of a chi-squared distribution: f(y)=1/(2^(n/2)*Γ(n/2))*y^(n/2-1)*exp(-y/2).

Next, we can use the same approach to derive the distribution of 2θ∑_(i=1)^n▒x_ι, where x_1,……,x_n are independent random variables with the same p.d.f. as X. Again, we define a new random variable, Y=2θ∑_(i=1)^n▒x_ι, and substitute it into the p.d.f. for X. This gives us the p.d.f. for Y as 1/sqrt(2π)*y^(-1/2)*exp(-y/2). This is the p.d.f. of a chi-squared distribution with n degrees of freedom, as can be seen by comparing it to the general form of a chi-squared distribution with n degrees of freedom: f(y)=1/(2^(n/2)*Γ(n/2))*y^(n/2-1)*exp(-y/2).

In order to recognize the degrees of freedom of this distribution, we need to consider the number of independent random variables involved in the sum. In this case, there are n random variables (x_1,……,x_n), so the resulting distribution has n