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Here is an image of the conversation which I had with my Prof. (He's the one in violet and myself in orange)

The topic was random variables and other probability related definitions. I tried to answer what all I knew but it was very unsatisfactory according to him. Can anyone please identify the blunders I made?

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I don't really understand the discussion about functionals. A functional $\phi$ on a (vector) space $X$ is a mapping $\phi : X \rightarrow \mathbb{R}$ (or $\mathbb{C}$). A linear functional is a functional such that $\phi(\alpha x + y) = \alpha\phi(x) + \phi(y)$ for all $\alpha \in \mathbb{R}$ and $x, y \in X$.

A distribution function $F$ is a right-continuous, non-decreasing function such that $P(X \leq x) = F(x)$. In particular, this implies $\lim_{x \rightarrow -\infty} F(x) = 0$ and $\lim_{x \rightarrow \infty} = 1$.

A probability density function $f$ is not a distribution function (you seem to understand this, though). It doesn't always exist, but when it does, we have $F(x) = \int_{-\infty}^x f(y)dy.$

You might want to consider the following: the sample space vs. state space.

The sample space, which I will denote $\Omega$, is the space of all possible events. Without getting into measure-theoretic details, we often talk about the probability of an event $A$ and we denote it by $P(A)$. The individual points in $\Omega$ are usually denoted $\omega$, and these are your $H$ or $T$. If you flip two coins, then you have 4 $\omega$s: $\omega_1 = \{H, H\}$, $\omega_2 = \{H, T\}, \omega_3 = \{T, H\}, \omega_4 = \{T, T\}$. Then, e.g., $A = \{\text{get heads at least once}\} = \{\omega_1, \omega_2, \omega_3\}$.

It is often convenient to express these events as random variables, which are functions from $\Omega$ to a state space. I will denote this space $S$. It is often $\mathbb{R}$, but it doesn't have to be. [Aside: It could be a function space, for instance: for every $\omega \in \Omega$, $X_t(\omega)$ is a function (of $t$ in this case). It could be a vector or any other number of things.]

Here's the thing your prof. was probably getting at: probabilists often say you can "forget" $\Omega$; all you need is the distribution of $X$ (for a suitable $X$). For instance, say all I care about is the number of heads. Consider

$$F(x) = \begin{cases} 0 & x < 0 \\ 1/4 & 0 \leq x < 1 \\ 3/4 & 1 \leq x < 2 \\ 1 & x \geq 2.\end{cases}$$

This is the distribution function of the random variable $X : \Omega \rightarrow S$ such that $X(\omega) = \{\text{number of heads in the event } \{\omega\}\}$.

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  • $\begingroup$ Shouldn't it be $0,\frac { 1 }{ 4 } ,\frac { 3 }{ 4 } ,0$? $\endgroup$ – Hyperbola Jul 13 '14 at 18:16
  • $\begingroup$ No. Remember that $F(x)$ has to be $1$ as $x \rightarrow \infty$. The probability of no heads is 1/4; the probability of exactly one head is 1/2, so the probability of at most one head ($P(X \leq 1)$) is 1/4 + 1/2 = 3/4. The probability of at most two heads is 1, because you will never get heads more than twice when flipping two coins. $\endgroup$ – snar Jul 13 '14 at 18:19
  • $\begingroup$ But then, isn't the sum of a distribution always equal to $1$? $\endgroup$ – Hyperbola Jul 13 '14 at 18:22
  • $\begingroup$ You should think of a distribution as a "distribution of mass." The integral or sum of a density function is always 1, because $\lim_{x \rightarrow \infty} \int_{-\infty}^x f(y) dy = \lim_{x\rightarrow \infty} = 1$. $\endgroup$ – snar Jul 13 '14 at 20:57

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