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In the following link, Distribution of Sum of Brownian Motion and Integrated BM, it is said that

1- $W(t) + \int_0^T W(t) dt$ does not exist, why?

2- What topic should I read or learn to understand whatever "Saz" has commented under the question? What is $t_j^n := T \cdot j/n dt$? Why do we define it? What is $X^n$? Why is it defined at all? I am very lost and I don't know what topic to study to understand those.

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Tayebe
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  • The reason is simply that "$t$" is used twice: as a variable (in the first term) and as a variable of integration (in the second term) $$W(\color{red}{t}) + \int_0^T W(t) , d\color{red}{t}$$ ... and this doesn't make sense. However, as Did pointed out, e.g. the expression $$W(\color{blue}{T}) + \int_0^T W(t) , dt$$ is well-defined. 2. $t_j^n = T j/n , dt$....? I didn't write this. I defined $t_j^n := T \cdot j/n$ for $j=0,\ldots,n$ and each $t_j^n$ is obviously just a non-negative number. (And I did define $X^n$ in my comment, so please be more speicific.....)
    • – saz Dec 18 '16 at 07:19
  • Thanks for your reply. My question, which may have been put badly, is actually very basic. Firstly, what is the nature of $X^n$? is it a random variable? Also, I fail to understand how defining $X^n$ helps us answer bcf's question? I believe the question is: the sum of two independent normally distributed random variables is always normally distributed too. In this question, there is no information about $W(t) $ and $\int_0^T W(t) dt$ being independent and in fact they are dependent. – Tayebe Dec 18 '16 at 10:29
  • This is not about independence. If a random vector $Y:=(Y_1,\ldots,Y_n)$ is Gaussian, then any linear combination $\sum_{j=1}^n a_j Y_j$ is Gaussian (here $a_j \in \mathbb{R}$ are (deterministic) constants). – saz Dec 18 '16 at 11:56