How do I define probability space $(Omega, mathcal F, mathbb{P})$ for continuous random variable?

I need to mathematically define the probability space $(\Omega, \mathcal F, \mathbb P)$ of continuous random variable $X$. I also need to define the continuous random variable $X$ itself. Problem is... I don't really know how.

It is known that $X$ has the following probability density function $f_X: \mathbb{R} \longrightarrow \left[0, \frac{4}{9} \right]$:

$$f_X(x) = \begin{cases} \begin{align*} &\frac{1}{9}\big(3 + 2x - x^2 \big) \; &: 0
\leq x \leq 3 \\ &0 \; \; &: x < 0 \; \lor \; x > 3 \end{align*}\end{cases}$$

and its plot:

Also, the cumulative distribution function of $X$ is $F_X: \; \mathbb{R} \longrightarrow \left[0,1\right]$ and is defined as:

$$F_X(x) = \begin{cases} \begin{align*} &0 \; \; &: x < 0 \\ &\frac{1}{9} \Big(3x + x^2 - \frac{1}{3}x^3 \Big) \; \; &: x \geq 0 \; \land \; x \leq 3 \\ &1 \; \; &: x > 3 \end{align*}\end{cases}$$

and its plot:

(please see this thread where I calculated CDF for reference)

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I suppose:

$$X: \Omega \longrightarrow \mathbb{R}$$

and sample space:

$$\Omega = \mathbb{R}$$

How can I define $\mathcal F$ and $\mathbb{P}$, that are the quantities of probability space $(\Omega, \mathcal F, \mathbb{P})$? I was thinking:

$$\mathbb{P} : \mathcal F \longrightarrow \left[0, 1\right] \; \land \; \mathbb{P}(\Omega) = 1$$

I am jumping into statistics/probability and I am lacking the theoretical knowledge. Truth be speaking, the wikipedia definition of probability space for continuous random variable is too difficult to grasp for me.

Thanks!

Answer

It is a bit weird to ask for a probability space if the probability distribution is already there and is completely at hand. So I think this is just some theoretical question to test you. After all students in probability theory must be able to place the "probability things" they meet in the confidential context of a probability space.

In such case the easyest way is the following.

Just take $(\Omega=\mathbb R,\mathcal F=\mathcal B(\mathbb R),\mathbb P$) as probability space where $\mathcal B(\mathbb R)$ denotes the $\sigma$-algebra of Borel subsets of $\mathbb R$ and where probability measure $\mathbb P$ is prescribed by: $$B\mapsto\int_Bf_X(x)\;dx$$

Then as random variable $X:\Omega\to\mathbb R$ you can take the identity on $\mathbb R$.

The random variable induces a distribution denoted as $\mathbb P_X$ that is characterized by $$\mathbb P_X(B)=\mathbb P(X\in B)=\mathbb P(X^{-1}(B))\text{ for every }B\in\mathcal B(\mathbb R)$$

Now observe that - because $X$ is the identity - we have $X^{-1}(B)=B$ so that we end up with:$$\mathbb P_X(B)=\int_Bf_X(x)\;dx\text{ for every }B\in\mathcal B(\mathbb R)$$as it should. Actually in this special construction we have:$$(\Omega,\mathcal F,\mathbb P)=(\mathbb R,\mathcal B(\mathbb R),\mathbb P_X)\text{ together with }X:\Omega\to\mathbb R\text{ prescribed by }\omega\mapsto\omega$$

Above we created a probability space together with a measurable function $\Omega\to\mathbb R$ such that the induced distribution on $(\mathbb R,\mathcal B(\mathbb R))$ is the one that is described in your question.

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PS: As soon as you are well informed about probability spaces then in a certain sense you can forget about them again. See this question to gain some understanding about what I mean to say.

radicals - How to prove that $sqrt 3$ is an irrational number?

I know how to prove $\sqrt 2$ is an irrational number. Who can tell me that why $\sqrt 3$ is a an irrational number?

probability - The density of a random variable $X$ is $f(x)$ proportional to $x^{-1/2}$ , what is the mean of $X$?

The density of a random variable $X$ is

$f(x)$ proportional to $x^{-1/2}$ for $x \in [0,1]$$

and $f(x) = 0$ for $x \notin [0,1]$. Then, the mean of $X$ is

1. $\frac 12$


2. $\frac 1{\sqrt2}$


3. $\frac 13$


4. $\frac 14$


5. None of the above is correct.

By the formula $\int_{0}^1 x\times x^{-1/2} dx $ (the formula of the expectation of continuous r.v.), I calculate the answer is $\frac 23$, but what is the meaning of the words proportional to? If I multiply some number, option 1-4 are both correct, so the answer is Option 5?

Answer

Using the fact that the density must integrate to one over $[0,1]$, then proportional means
$$1 = \int_0^1cf(x)\,dx = \int_0^1\frac{c}{\sqrt x}\, dx = c\left[2\sqrt x\right]_0^1 = 2c.$$
Thus $c = 1/2$.

If you now try to compute the expectation, you will find
$$\int_0^1x\cdot \frac{1/2}{\sqrt x} = \frac{1}{3},$$
which is option 3.

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Given the following series

$$\sum_{k=0}^\infty \frac{\sin 2k}{1+2^k}$$

I'm supposed to determine whether it converges or diverges. Am I supposed to use the comparison test for this? My guess would be to compare it to $\frac{1}{2^k}$ and since that is a geometric series that converges, my original series would converge as well. I'm not all too familiar with comparing series that have trig functions in them. Hope I'm going in the right direction

Thanks

Answer

You have the right idea, but you need to do a little more, since some of the terms are negative. Use your idea and the fact that $|\sin x|\le 1$ for all $x$ to show that

$$\sum_{k\ge 0}\frac{\sin 2k}{1+2^k}$$

is absolutely convergent, i.e., that

$$\sum_{k\ge 0}\left|\frac{\sin 2k}{1+2^k}\right|$$

converges.

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I am learning about the derivative function of $\frac{d}{dx}[\sin(x)] = \cos(x)$.

The proof stated: From $\lim_{x \to 0} \frac{\sin(x)}{x} = 1$...

I realized I don't know why, so I wanted to learn why part is true first before moving on. But unfortunately I don't have the complete note for this proof.

1. It started with a unit circle, and then drew a triangle at $(1, \tan(\theta))$


2. It show the area of the big triangle is $\frac{\tan\theta}{2}$


3. It show the area is greater than the sector, which is $\frac{\theta}{2}$
   Here is my question, how does this "section" of the circle equal to $\frac{\theta}{2}$? (It looks like a pizza slice).



4. From there, it stated the area of the smaller triangle is $\frac{\sin(\theta)}{2}$. I understand this part. Since the area of the triangle is $\frac{1}{2}(\text{base} \times \text{height})$.




5. Then they multiply each expression by $\frac{2}{\sin(\theta){}}$ to get
   $\frac{1}{\cos(\theta)} \ge \frac{\theta}{\sin(\theta)} \ge 1$

And the incomplete notes ended here, I am not sure how the teacher go the conclusion $\lim_{x \to 0} \frac{\sin(x)}{x} = 1$. I thought it might be something to do with reversing the inequality... Is the answer obvious from this point? And how does step #3 calculation works?

Answer

Draw the circle of radius $1$ centered at $(0,0)$ in the Cartesian plane.

Let $\theta$ be the length of the arc from $(1,0)$ to a point on the circle. The radian measure of the corresponding angle is $\theta$ and the height of the endpoint of the arc above the coordinate axis is $\sin\theta$.

Now look at what happens when $\theta$ is infinitesimally small. The length of the arc is $\theta$ and the height is also $\theta$, since that infinitely small part of the circle looks like a vertical line (you're looking at the neighborhood of $(1,0)$ under a microscope).

Since $\theta$ and $\sin\theta$ are the same when $\theta$ is infinitesimally small, it follows that $\dfrac{\sin\theta}\theta=1$ when $\theta$ is infinitesimally small.

That is how Leonhard Euler viewed the matter in the 18th century.

Why does the sector of the circle have area $\theta/2$?

The whole circle has area $\pi r^2=\pi 1^2 = \pi$. The fraction of the circle in the sector is
$$
\frac{\text{arc}}{\text{circumference}} = \frac{\theta}{2\pi}.
$$
So the area is
$$
\frac \theta {2\pi}\cdot \pi = \frac\theta2.

$$

operations on real functions

If $f$ and $g$ are two real functions such that domain of $f$ is $D_1$ and domain of $g$ is $D_2$ both being subsets of $\mathbb R$ . My book says that the function $f+g$ will have the domain ($D_1 \cap D_2$). Why is this? And if $f$ and $g$ have co-domain $C_1$ and $C_2$ respectively such that $C_1$ and $C_2$ are subsets of set of real numbers then what will be the co-domain of $f+g$?

calculus - One sided limit with $e^x$ and $sin(x)$

Assuming that $e^{x} \to e^{a}$ and $\sin(x) \to \sin(a)$

evaluate the limit as $$\lim_{x \to 0^{+}} \frac{e^{x^2+2x-1}}{\sin(x)}$$

To me, based on the first sentence. I could just plug in $a$ and get my limit which does not exist at the point $x=0$ However the back of the book says infinity so I am at a loss as to how to prove this.

This is a real analysis class but this is isn't one of the problems requiring us to prove with definitions but it would be helpful to learn the algebra and the theory. I don't understand as to how I can apply the definition of a function converging to infinity to prove this. Plus I don't see how that could help? How am I supposed to know it goes to infinity

Answer

When $x$ is close enough to $0$, then $-1.1 so that
$$\exp(-1.1)<\exp(x^2+2x-1)$$

and hence the function in your limit is bounded from below by $\frac{\exp(-1.1)}{\sin x}$ which blows up at $x=0$ since $\sin0=0$.