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Is the sum of two binomial distributions a binomial distribution?
In probability theory and statistics, the sum of independent binomial random variables is itself a binomial random variable if all the component variables share the same success probability. If success probabilities differ, the probability distribution of the sum is not binomial.
How do you combine two binomials?
Use the FOIL method for multiplying two binomials
- Multiply the First terms.
- Multiply the Outer terms.
- Multiply the Inner terms.
- Multiply the Last terms.
- Combine like terms, when possible.
What does binomial CDF find?
The binomial cumulative distribution function lets you obtain the probability of observing less than or equal to x successes in n trials, with the probability p of success on a single trial.
How to calculate CDF of sum of independent random variables?
F Z ( z) = P ( Z ≤ z) = P ( X + Y ≤ z) = ∫ R ∫ − ∞ z − x f X, Y ( x, y) d y d x = ∫ R F Y ( z − x) f ( x) d x. where ∗ stands for convolution. Thanks for contributing an answer to Mathematics Stack Exchange! Please be sure to answer the question. Provide details and share your research! But avoid …
Which is the sum of two independent binomial variables?
Since X, Y independent, φX + Y(t) = φX(t)φY(t) = (1 − p + peit)n(1 − p + peit)m = (1 − p + peit)n + m. By uniqueness, we get X + Y ∼ Bin (n + m, p). We can prove this using Moment generating function as follows if someone is not comfortable with characterstic functions as answered above:
Is there a way to express the cumulative distribution function?
It’s easy to check that the moment generating function ϕ Z ( t) := E [ e i t Z] can be expressed as ϕ Z = ϕ X ⋅ ϕ Y. Is there a way to express the cumulative distribution function F Z ( z) := P ( Z ≤ z) using the cumulative distribution functions of X and Y ? Edit: note I don’t assume that X and Y have a density.
Is the sum of X and Y independent?
It is evident that X and Y are independent. Now realize that X + Y = B1 + ⋯ + Bn + m is binomially distributed with parameters n + m, p. This spares you any computations. Just compute. Suppose X ∼ Bin(n, p), Y ∼ Bin(m, p).