For the multinomial distribution:
Show that summation of the probability mass function across all possible values equals 1.
Show via summation of the probability mass function that the expected value of Y_i = n * p_i
The textbook that I'm using left the above as "exercises to the reader". I've wracked my brain and am completely stumped on how to get started.
To prove that the summation of the probability mass function (PMF) of a multinomial distribution across all possible values equals 1, we need to use the fact that the total probability of all possible outcomes in any probability distribution must equal 1.
Let's assume we have a multinomial distribution with k possible outcomes and probabilities (p₁, p₂,..., pₖ) for each outcome. The variable Y denotes the number of times each outcome occurs.
The PMF of a multinomial distribution is given by:
P(Y₁ = y₁, Y₂ = y₂,..., Yₖ = yₖ) = (n! / (y₁! * y₂! * ... * yₖ!)) * (p₁^y₁) * (p₂^y₂) * ... * (pₖ^yₖ)
Here, n is the total number of trials (i.e., n = y₁ + y₂ + ... + yₖ), and y₁, y₂,..., yₖ are the specific values of Y₁, Y₂,..., Yₖ, respectively.
To prove that the summation of the PMF across all possible values equals 1, we need to sum over all possible combinations of y₁, y₂,..., yₖ that satisfy the constraint n = y₁ + y₂ + ... + yₖ.
The sum can be written as follows:
∑∑...∑ P(Y₁ = y₁, Y₂ = y₂,..., Yₖ = yₖ) = 1
where the double summation represents the sum over all possible combinations of y₁, y₂,..., yₖ.
To evaluate this sum, we can use the concept of the binomial coefficient. The binomial coefficient can be calculated as:
C(n, r) = n! / (r! * (n-r)!)
where n! represents the factorial of n.
Since the denominator of the multinomial PMF includes factorials (y₁!, y₂!,..., yₖ!), we can factor out these binomial coefficients. The remaining terms in the numerator simplify to p₁^y₁ * p₂^y₂ * ... * pₖ^yₖ.
By applying the principle of the multinomial theorem and simplifying the expression, we can conclude that the summation of the PMF across all possible values equals 1.
Now, to show that the expected value of Yᵢ, denoted E(Yᵢ), is equal to n * pᵢ, we can use the properties of the expected value:
E(Yᵢ) = ∑ yᵢ * P(Y₁ = y₁, Y₂ = y₂,..., Yₖ = yₖ)
By substituting the PMF of the multinomial distribution, we have:
E(Yᵢ) = ∑ yᵢ * [(n! / (y₁! * y₂! * ... * yₖ!)) * (p₁^y₁) * (p₂^y₂) * ... * (pₖ^yₖ)]
Simplifying this expression will involve using the properties of the binomial coefficient, as well as algebraic manipulation. By following these steps, you should be able to derive the result that E(Yᵢ) = n * pᵢ.
It is important to note that these proofs involve advanced mathematical concepts and calculations. If you are struggling to solve these exercises on your own, it would be helpful to consult textbooks, online resources, or seek guidance from a teacher or tutor.