0
$\begingroup$

This is a mixture binomial distribution question. I know how to get the $\mu$ and $\sigma^2$ of the mixture, but I am not sure how to use it to get the probably of specific number.

Question:
States that $X_1=B(2,0.52), X_2=B(3,0.41), X_3=B(4,0.38)$ are binomial distributions with 43%, 36%, 21% users respectively. Find the probability that occurrence is more than 2. The question don't state whether the variables are independence of each other. I will assume this is a mixture distribution question.

Binomial distribution: https://en.wikipedia.org/wiki/Binomial_distribution

Reference for mixture distribution: https://en.wikipedia.org/wiki/Mixture_distribution

My solution:
$X_1 = B(2, 0.52), E(X_1)=1.04, E(X_1^2)=1.5808, Var(X_1)=0.4992$

$X_2 = B(3, 0.41), E(X_2)=1.23, E(X_2^2)=2.2386, Var(X_2)=0.7257$

$X_3 = B(4, 0.38), E(X_3)=1.52, E(X_3^2)=3.2528, Var(X_3)=0.9424$

I see S as a mixture distribution of $X_1, X_2, X_3$

$P(S=0) = 0.43P(X_1=0) + 0.36P(X_2=0) + 0.21P(X_3=0)$

$E(S) = 0.43E(X_1) + 0.36E(X_2) + 0.21E(X_3)=1.2092$

$E(S^2)=0.43E(X_1^2) + 0.36E(X_2^2) + 0.21E(X_3^2)=2.168728$

$Var(S)=E(S^2)-(E(S))^2=0.706563$

Stuck:
How do I get $P(S<=2)$? Do I do $P(\frac{S-1.2092}{\sqrt{0.706563}}<=\frac{2-1.2092}{\sqrt{0.706563}})$? Is normal distribution method of getting the probability correct?

Or should I do $P(S<=2) = 1-P(S>2)$ where $P(S>2)=0.43P(X_1>2)+0.36P(X_2>2)+0.21P(X_3>2)$

$\endgroup$
2
$\begingroup$

First $-$ your calculation of $\text{VAR}(S)$ is wrong $-$ the coefficients have to be squared when $S$ is a linear combination (assuming of course that $X_1,X_2,X_3$ are independent random variables). Second $-$ the normal approximation does not seem to be accurate with such small numbers as $2,3,4$. I would use "brute force" calculation $-$ $X_1$ can be 0,1,2 ; $X_2$ can be $0,1,2,3$ ; $X_3$ can be $0,1,2,3,4$. There are 60 possible results - see which ones result in $S\leq 2$ and add up their probabilities (each probability in the sum will be a product of three binomial probabilities). Looking at the numbers, it will be easier to calculate $P(S>2)$ and then subtract it from $1$. There are less combinations which will result in $S>2$.

$\endgroup$
  • $\begingroup$ I got $Var(S) = E(S^2)-(E(S))^2$, what do you mean by squaring the coefficients? $\endgroup$ – shawnngtq Sep 10 '17 at 1:48
1
$\begingroup$

Can you find the probability that $P(X_i ≤ 2)$ for each $i$? If so, it's straightforward to use the law of total probability to compute $P(S ≤ 2)$, similarly to how you propose to find $P(S > 2)$. It should be clear that the answer you get this way is exact. The normal approximation you propose doesn't make a lot of sense because $S$ is not defined as a sum of random variables.

$\endgroup$
  • $\begingroup$ I'm a little confused by your answer. As a mixture, the possible values taken by S will be all the possible values taken by the components (the $X_i$). So S will take values in $\{0,1,2,3,4\}$ ... i.e. always integer. Are you sure you understood the question? $\endgroup$ – Glen_b -Reinstate Monica Sep 10 '17 at 8:08
  • $\begingroup$ @Glen_b The question has been edited. It originally defined $S$ as $S = 0.43X_1 + 0.36X_2 + 0.21X_3$, rather than a conventional mixture distribution. $\endgroup$ – Kodiologist Sep 10 '17 at 14:17
  • $\begingroup$ Ah. I see. However, every version of the question opens with "This is a mixture binomial distribution question". The first version (the one up at the time you answered) does contain the line $S = 0.43X_1 + 0.36X_2 + 0.21X_3$ but in the context of that opening statement, one would have to be cautious about interpreting that at face value (since it would not be a mixture interpreted the way you did). $\endgroup$ – Glen_b -Reinstate Monica Sep 10 '17 at 16:41
  • $\begingroup$ @Glen_b I figured that the equation would be likelier to be copied from the textbook exactly, and hence would be more trustworthy, than the words. Looks like I guessed wrong. I'll edit my answer. $\endgroup$ – Kodiologist Sep 10 '17 at 17:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.