# How can we calculate the probability?

• MHB
• mathmari
In summary, the conversation discusses the calculation of the distribution function and probability function for a random variable. It includes a discussion of how to calculate probabilities for different values of the random variable, and the use of the Dirac delta function for non-differentiable points in the distribution function. The final steps of the calculation for a specific probability are also shown.
mathmari
Gold Member
MHB
Hey! :giggle:

The distribution of a random variable $X:\Omega\rightarrow \mathbb{R}$ is given by the probability function $p_X$ as follows:

(i) Calculate the distribution function $F_X$ of $X$.
(ii) Let $Y(\omega)=|X(\omega)|$ for all $\omega\in \Omega$. Determine $p_Y$. I have done the following :

(i) \begin{align*}&F_X(-4)=p_X(-4)=\frac{1}{4} \\ &F_X(0)=p_X(-4)+p_X(0)=\frac{1}{4}+\frac{1}{6}=\frac{5}{12} \\ &F_X(4)=p_X(-4)+p_X(0)+p_X(4)=\frac{1}{4}+\frac{1}{6}+\frac{1}{4}=\frac{5}{12}+\frac{1}{4}=\frac{2}{3} \\ &F_X(12)=p_X(-4)+p_X(0)+p_X(4)+p_X(12)=\frac{1}{4}+\frac{1}{6}+\frac{1}{4}+\frac{1}{3}=\frac{2}{3}+\frac{1}{3}=1\end{align*}

(ii) We have that $Y(\Omega)\in \{0, 4, 12\}$, right? But how can we calculate the probability? :unsure:

mathmari said:
(ii) We have that $Y(\Omega)\in \{0, 4, 12\}$, right? But how can we calculate the probability?

Hey mathmari!

Don't we have for $y\ge 0$ that $p_Y(y)=P(Y(\omega)=y)=P(|X(\omega)|=y)=P(X(\omega)=y \lor X(\omega)=-y)$?

Btw, we have for $\omega\in \Omega$ that $Y(\omega)\in \{0, 4, 12\}$.
However $Y(\Omega)= \{0, 4, 12\}$. (Nerd)

Klaas van Aarsen said:
Don't we have for $y\ge 0$ that $p_Y(y)=P(Y(\omega)=y)=P(|X(\omega)|=y)=P(X(\omega)=y \lor X(\omega)=-y)$?

Btw, we have for $\omega\in \Omega$ that $Y(\omega)\in \{0, 4, 12\}$.
However $Y(\Omega)= \{0, 4, 12\}$. (Nerd)

So we get $$p_Y(0)=\frac{1}{6}, \ p_Y(4)=\frac{1}{4}, \ p_Y(12)=\frac{1}{3}$$ right? :unsure:

mathmari said:
So we get $$p_Y(0)=\frac{1}{6}, \ p_Y(4)=\frac{1}{4}, \ p_Y(12)=\frac{1}{3}$$ right?

They should sum up to 1...
It seems we missed a chance somewhere. (Worried)

Klaas van Aarsen said:
They should sum up to 1...
It seems we missed a chance somewhere. (Worried)

Do we maybe have the following :
$$p_Y(4)=P(X(\omega)=4 \lor X(\omega)=-4)=P(X(\omega)=4) + P(X(\omega)=-4)=\frac{1}{4}+\frac{1}{4}=\frac{2}{4}=\frac{1}{2}$$
:unsure:

mathmari said:
Do we maybe have the following :
$$p_Y(4)=P(X(\omega)=4 \lor X(\omega)=-4)=P(X(\omega)=4) + P(X(\omega)=-4)=\frac{1}{4}+\frac{1}{4}=\frac{2}{4}=\frac{1}{2}$$
Yep. (Nod)

Klaas van Aarsen said:
Yep. (Nod)

So for all probabilities we have :
\begin{align*}p_Y(0)&=P(Y(\omega)=0)=P(|X(\omega)|=0)=P(X(\omega)=0 \lor X(\omega)=-0)=P(X(\omega)=0)\\ & = \frac{1}{6} \\
p_Y(4)&=P(Y(\omega)=4)=P(|X(\omega)|=4)=P(X(\omega)=4 \lor X(\omega)=-4)=P(X(\omega)=4) + P(X(\omega)=-4)\\ & =\frac{1}{4}+\frac{1}{4}=\frac{2}{4}=\frac{1}{2} \\
p_Y(12)&=P(Y(\omega)=12)=P(|X(\omega)|=12)=P(X(\omega)=12 \lor X(\omega)=-12)\\ & =P(X(\omega)=12) + P(X(\omega)=-12)=\frac{1}{3}+0=\frac{1}{3}\end{align*}
right? :unsure:

Yep. (Nod)

And they sum up to 1 now as they should.(Emo)

Great! (Clapping)At the next subquestion we have :
Let $Z : \Omega \rightarrow \mathbb{R}$ be a random variable with distribution function $F_Z : \mathbb{R} \rightarrow [0, 1]$,

(i) Calculate $p_Z$.
(ii) Calculate $P[2 \leq Z < 3]$.I have done the following :

(i) \begin{align*}&p_Z(1)=F_Z(1)=\frac{2}{5} \\ &p_Z(2)=F_Z(2)-F_Z(1)=\frac{3}{5}-\frac{2}{5}=\frac{1}{5} \\ &p_Z\left (\frac{5}{2}\right )=F_Z\left (\frac{5}{2}\right )-F_Z(2)=\frac{9}{10}-\frac{3}{5}=\frac{9}{10}-\frac{6}{10}=\frac{3}{10} \\ &p_Z(3)=F_Z(3)-F_Z\left (\frac{5}{2}\right )=1-\frac{9}{10}=\frac{1}{10}\end{align*}

(ii) \begin{equation*}P[2\leq Z<3]=p_Z(2)=F_Z(2)-F_Z(1)=\frac{3}{5}-\frac{2}{5}=\frac{1}{5}\end{equation*} Is everything correct and complete? :unsure:

Last edited by a moderator:
$Z$ is a continuous random variable now, which means we need to specify $p_Z$ for every real number.
What is for instance $p_Z(1.5)$? (Wondering)

Klaas van Aarsen said:
$Z$ is a continuous random variable now, which means we need to specify $p_Z$ for every real number.
What is for instance $p_Z(1.5)$? (Wondering)

Do you mean that we have to write :
\begin{align*}&p_Z(x)=F_Z(1)=\frac{2}{5} \ \text{ for } 1\leq x<2 \\ &p_Z(x)=F_Z(2)-F_Z(1)=\frac{3}{5}-\frac{2}{5}=\frac{1}{5} \ \text{ for } 2\leq x<\frac{5}{2} \\ &p_Z\left (x\right )=F_Z\left (\frac{5}{2}\right )-F_Z(2)=\frac{9}{10}-\frac{3}{5}=\frac{9}{10}-\frac{6}{10}=\frac{3}{10} \ \text{ for } \frac{5}{2}\leq x<3 \\ &p_Z(x)=F_Z(3)-F_Z\left (\frac{5}{2}\right )=1-\frac{9}{10}=\frac{1}{10} \ \text{ for } x>3 \end{align*} Or what do we do in this case? :unsure:

mathmari said:
Do you mean that we have to write :
\begin{align*}&p_Z(x)=F_Z(1)=\frac{2}{5} \ \text{ for } 1\leq x<2 \\ &p_Z(x)=F_Z(2)-F_Z(1)=\frac{3}{5}-\frac{2}{5}=\frac{1}{5} \ \text{ for } 2\leq x<\frac{5}{2} \\ &p_Z\left (x\right )=F_Z\left (\frac{5}{2}\right )-F_Z(2)=\frac{9}{10}-\frac{3}{5}=\frac{9}{10}-\frac{6}{10}=\frac{3}{10} \ \text{ for } \frac{5}{2}\leq x<3 \\ &p_Z(x)=F_Z(3)-F_Z\left (\frac{5}{2}\right )=1-\frac{9}{10}=\frac{1}{10} \ \text{ for } x>3 \end{align*} Or what do we do in this case?

It's a bit strange really.
We have $F_Z(z) = \int_{-\infty}^z p_Z(x)\,dx \implies p_Z(x) = F_Z'(x)$.
That means that $p_Z(x)=0$ for $1<x<2$.
However, $F_Z(x)$ is not differentiable at $x=1$, so we would need to introduce the Dirac delta function to describe $p_Z(x)$ there.

Last edited:
Klaas van Aarsen said:
It's a bit strange really.
We have $F_Z(z) = \int_{-\infty}^z p_Z(x)\,dx \implies p_Z(x) = F_Z'(x)$.
That means that $p_Z(x)=0$ for $1<x<2$.
However, $F_Z(x)$ is not differentiable at $x=1$, so we would need to introduce that Dirac delta function to describe $p_Z(x)$ there.

Is this related to the below example?

:unsure:

mathmari said:
Is this related to the below example?
Yes. That example shows $p_X$ as a discrete function and $F_X$ as a continuous function.
If we follow that example, then I think that what you wrote before for $p_Z$ as a discrete function is correct.

Klaas van Aarsen said:
Yes. That example shows $p_X$ as a discrete function and $F_X$ as a continuous function.
If we follow that example, then I think that what you wrote before for $p_Z$ as a discrete function is correct.

Do you mean what I wrote in post #9 or in post #11 ? :unsure:

mathmari said:
(ii) \begin{equation*}P[2\leq Z<3]=p_Z(2)=F_Z(2)-F_Z(1)=\frac{3}{5}-\frac{2}{5}=\frac{1}{5}\end{equation*}

I wrote all steps analytically as follows :
\begin{align*}P[2\leq Z<3]&=P[\{Z<3\}\setminus \{Z<2\}]=P[Z<3]-P[Z<2] \\ &=P[\{Z\leq 3\}\setminus \{Z=3\}]-P[\{Z\leq 2\}\setminus \{Z=2\}]\\ & =\left (P[Z\leq 3]-P[Z=3]\right )-\left (P[Z\leq 2]-P[Z=2]\right ) \\ &=\left (F_Z(3)-P[Z=3]\right )-\left (F_Z(2)-P[Z=2]\right )\\ &=\left (F_Z(3)-p_Z(3)\right )-\left (F_Z(2)-p_Z(2)\right )\\ &=\left (1-\frac{1}{10}\right )-\left (\frac{3}{5}-\frac{1}{5}\right )\\ &=1-\frac{1}{10}-\frac{3}{5}+\frac{1}{5}\\ & =\frac{1}{2}
\end{align*}
Are all steps correct? :unsure:

mathmari said:
Do you mean what I wrote in post #9 or in post #11 ?
Post #9.

mathmari said:
I wrote all steps analytically as follows :
\begin{align*}P[2\leq Z<3]&=P[\{Z<3\}\setminus \{Z<2\}]=P[Z<3]-P[Z<2] \\ &=P[\{Z\leq 3\}\setminus \{Z=3\}]-P[\{Z\leq 2\}\setminus \{Z=2\}]\\ & =\left (P[Z\leq 3]-P[Z=3]\right )-\left (P[Z\leq 2]-P[Z=2]\right ) \\ &=\left (F_Z(3)-P[Z=3]\right )-\left (F_Z(2)-P[Z=2]\right )\\ &=\left (F_Z(3)-p_Z(3)\right )-\left (F_Z(2)-p_Z(2)\right )\\ &=\left (1-\frac{1}{10}\right )-\left (\frac{3}{5}-\frac{1}{5}\right )\\ &=1-\frac{1}{10}-\frac{3}{5}+\frac{1}{5}\\ & =\frac{1}{2}
\end{align*}
Are all steps correct?
Correct although I believe it suffices to write:
\begin{align*}P[2\leq Z<3]= P(Z<3)-P(Z<2) =\lim_{x\to 3^-} F_Z(x) - \lim_{x\to 2^-} F_Z(x) = \frac 9{10}-\frac 25 =\frac{1}{2}
\end{align*}

Klaas van Aarsen said:
Post #9. Correct although I believe it suffices to write:
\begin{align*}P[2\leq Z<3]= P(Z<3)-P(Z<2) =\lim_{x\to 3^-} F_Z(x) - \lim_{x\to 2^-} F_Z(x) = \frac 9{10}-\frac 25 =\frac{1}{2}
\end{align*}

Ah ok! Thank you very much! (Sun)

## 1. What is probability and why is it important?

Probability is the measure of the likelihood that an event will occur. It is important because it helps us make decisions and predictions based on data and can inform us about the likelihood of certain outcomes.

## 2. How can we represent probability mathematically?

Probability can be represented as a number between 0 and 1, where 0 indicates impossibility and 1 indicates certainty. It can also be represented as a fraction, decimal, or percentage.

## 3. What are the different types of probability?

The three main types of probability are theoretical, experimental, and subjective. Theoretical probability is based on mathematical principles and assumptions, experimental probability is based on data collected from experiments, and subjective probability is based on personal beliefs and opinions.

## 4. How can we calculate the probability of independent events?

The probability of independent events can be calculated by multiplying the individual probabilities of each event. For example, if the probability of event A is 0.5 and the probability of event B is 0.3, the probability of both events occurring is 0.5 x 0.3 = 0.15.

## 5. How can we use probability to make predictions?

We can use probability to make predictions by analyzing data and calculating the likelihood of certain outcomes. By understanding the probability of different events, we can make informed decisions and predictions about the future.

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