find the probability mass function of a random variable X

Question

Here's a screenshot of the solution/explanation in my book, but I don't get it!:

1) How do you read "p(i) = cλ^i/i!" ? What do the symbols mean?

2) How did they derive the 2nd equation (infinity, i goes to zero, the summation of p(i) = 1)

3) In the 3rd equation, why did they isolate 'c' outside the summation?

4) in the 4th equation, why is e^x = the summation of x^i/i!, infinity, i goes to 0? On what basis did they come to this conclusion?

5) How did they simplify and find the 5th equation (ce^λ = 1 or c = e^-λ)

6) How did they solve to find (a) & (b)? Thank you!!!!

Answer 1

This is a long list of questions, so here is some answers.

1) $p(i) = \cdots$ means $P(X = i)$, the probability that $X$ takes value $i$.

2) That is the result of the second and third axioms of probability : $$1 = P(\Omega) = P(\bigcup_{i=0}^\infty \{X = i\}) = \sum_{i=0}^\infty P(\{X = i\})$$

3) In order to compute $c$. It is a constant and comes out of the sum.

4) Taylor expansion.

5) This is solving a linear equation in one unknown. Also $1/e^{\lambda} = e^{-\lambda}$.

6) (a) is by definition, see answer to 1. (b)is using properties of probability measures (i.e., consequences of the axioms). $\{X > 2\}$ is the complement of $\{X \le 2\}$ hence their probabilities sum to $1$. $\{X \le 2\}$ is the union of two disjoint events $\{X = 0\}$ and $\{X = 1\}$, so its probability is the sum of those two.

Answer 2

Ok. Starting with the first. $p(i)$ can be understood as $P(X=i)$. Note that this random variable takes only countable number of values. $i= 0, 1, 2, ...$. For the second part, they sum all thr probabilities. It should be 1. It can be understood like this. Your random variable $\textbf{must}$ take one of the values $0, 1, 2, ...$. It can't avoid taking one. So for sure it takes one of that. "For sure" can be tranlslated to be of probability one. One the other hand, it can take at time only one of the value. So the sum of the probabilities should be one. So for the second one they take the sum over all possible values it can take. See that multiplication of $c$ is everywhere, so it is regardless of what's your variables value. In other word it doesn't depend on $i$, so it can be taken out (as multiplication for the sum. It can be stated as something close to this $ca+cb= c(a+b)$). About your forth point. It is characterization of $e$. It is famous formula. About fifth equation. Just multipled both side with $e^{-\lambda}$. (and having that $e^{\lambda}e^{-\lambda}$= 1). About the sixth point. Finding $c$ in the form of soe function of $\lambda$, you can explicitely calculate, what's the probability of $X=i$, I mean $P(X=i)= p(i)$. Just take to be $i=0$, for the first case. For the second case however the following technique is used. If your random variable takes value more than $2$, this is the same as claiming that yuor random variable haven't taken values $0, 1, \text{and}, 2$. Knowing that the probability of whole is one, it mean you should just substract the probabilities $p(0), p(1), p(2)$, from $1$.

Answer 3

1) $p$ is the probability mass function, $i$ is the argument of the probability mass function (i.e. a positive integer), $p(i)$ is the probability mass function evaluation at the argument $i$, $c$ is a constant chosen so that $\sum_{i=0}^{\infty} p(i)=1$, $\lambda$ is a real number i.e. a parameter of the probability mass function, $\lambda^i$ is the number $\lambda$ raised to the $i$th power, $i!$ is the factorial of $i$ which equals $(i)(i-1)\dots (2)(1)$.

2) $\sum_{i=0}^{\infty} p(i) =1$ by the law of total probability (i.e. the sum of all values of a probability mass function must be $1$ by definition of probability mass function).

$p(i)=\mathbb{P}(X=i)$ by definition, and since $X$ has to equal some positive integer (i.e. with probability 1), the sum of all probabilities that it is equals a positive integer $\sum_{i=0}^{\infty} p(i)=\sum_{i=0}^{\infty} \mathbb{P}(X=i)$ (which is the same thing as the probability that $X$ equals some positive integer $\mathbb{P}(X=i \text{ for some }i\in\{0,1,2,3,\dots\})$ by the law of total probability) is equal to $1$.

3) This follows from the distributive property.

4) This is just the Taylor series for the exponential function, i.e. it is an equation whose proof one learns in the second semester of calculus typically. In other words, this is essentially the definition of $e^x$.

5) Because $e^x = \sum_{i=0}^{\infty} \frac{x^i}{i!}$ by definition, substituting $x=\lambda$ into the definition gives: $e^{\lambda} = \sum_{i=0}^{\infty} \frac{\lambda^i}{i!}$. Using the equation $c c \sum_{i=0}^{\infty} \frac{\lambda^i}{i!}=1$, this implies immediately that $c e^{\lambda} =1 \implies c = \frac{1}{e^{\lambda}} = e^{-\lambda}$.

6) This again just the definition of $p(i)=\mathbb{P}(X=i)$. For $\mathbb{P}(X>2)$ note that this equals $\mathbb{P}(X = \text{anything}) - \mathbb{P}(X = 0\text{ or }1\text{ or }2)= 1 -p(0)-p(1)-p(2)$.

Answer 4

1) How do you read "p(i) = cλ^i/i!" ? What do the symbols mean?

"The probability for $X$ to realise value $i$ equals $c$ times lambda ($\lambda$) to the power of $i$ divided by $i$ factorial"

$c$ and $\lambda$ are constants.   The parameters of the system.

2) How did they derive the 2nd equation (infinity, i goes to zero, the summation of p(i) = 1)

The probability that $X$ can realise any one of the supported values is $1$.   By definition of what is a probability mass function.   Since $X$ will certainly be one of any of the natural numers, then:

$$\sum_{i=0}^\infty p(i)=1$$

3) In the 3rd equation, why did they isolate 'c' outside the summation?

It's a constant and multiplication distributes over addition. That is, $ca+cb = c(a+b)$, and by induction:

$$\sum_{i=0}^\infty c\,a_i = c \sum_{i=1}^\infty a_i$$

4) in the 4th equation, why is $e^x =$ the summation of $x^i/i!$, infinity, i goes to 0? On what basis did they come to this conclusion?

It's the Taylor series expansion of the $x$ exponential of Euler's number around zero, a topic you really should have covered.

$$e^x = \sum_{i=0}^\infty \dfrac{x^i}{i!}$$

For now, just accept that it is a well known result that, when you study (or revise) Taylor series, you can derive for yourself.

5) How did they simplify and find the 5th equation (ce^λ = 1 or c = e^-λ)

It's just substitution of the above, then algebraic rearrangement.   If $ca=1$ then $c=a^{-1}$.

$$\begin{align}c\sum_{i=0}^\infty \dfrac{\lambda^i}{i!} =1 &\iff c~e^\lambda=1\\ & \iff c= \frac 1{e^\lambda} \\ & \iff c = e^{-\lambda}\end{align}$$

6) How did they solve to find (a) & (b)? Thank you!!!!

Because $\mathsf P(X=i)=p(i)$, then

(a) $\mathsf P(X=0) = p(0)$

(b) $\mathsf P(X>2) = 1-p(0)-p(1)-p(2)$

Then apply the formula, $p(i) = \dfrac{e^{-\lambda}\lambda^i}{i!}\quad\big[i\in\Bbb N\big]$