2
$\begingroup$

There are 3 types of flowers that can grow from planting a seed. $$P(\text{Daisy}) = \theta_1$$ $$P(\text{Rose}) = (1-\theta_1)\theta_2$$ $$P(\text{Sunflower}) = (1-\theta_1)(1-\theta_2)$$

the total number of flowers at the end is $n.$ If $X=(X_1, X_2, X_3)$ is the number of daisies, roses and sunflowers respectively, what is the probability mass function of $X$?

Edit: The answer below answered my question about the distribution. How can I find the max likelihood estimators for $\theta_1$ and $\theta_2$? Thank you so much!

$\endgroup$
3
  • $\begingroup$ This is a multinomial distribution. $\qquad$ $\endgroup$ Commented Mar 5, 2020 at 19:16
  • $\begingroup$ You can ask a new question for the mle of the multinomial distribution with $p_1=..., p_2=..., p_3=...$ That´s no problem. $\endgroup$ Commented Mar 5, 2020 at 20:09
  • $\begingroup$ There are severals answers according to the mle here. This link should help. $\endgroup$ Commented Mar 5, 2020 at 20:19

1 Answer 1

0
$\begingroup$

First of all, you need to confirm the multinomial distribution condition - constrain the probability to space [0,1] and the sum of the probabilities not more than one (i.e., $\sum\limits_{i=1}^3 p_i= 1$).

Hence, if P(Daisy) + P(Rose) + P(Sunflower) = 1 then the responses ($X$) are from multinomial distribution.

Thus the equation, which you have to check, is $$\theta_1+(1-\theta_1)\theta_2+(1-\theta_1)(1-\theta_2)=1 \quad \dots (1)$$

Is this true, if $\theta_1,\theta_2 \in \mathbb R$ ?

If the equation (1) above holds then indeed $X$ is a multinomial distributed random variable with $p_1=\theta_1, p_2=(1-\theta_1)\theta_2, p_3=(1-\theta_1)(1-\theta_2)$ and the pmf

$$f_{X}(x_1,x_2,x_3) = \displaystyle {n! \over x_1!\cdot x_2! \cdot x_3!} \cdot p_1^{x_1}\cdot p_2^{x_2}\cdot p_3^{x_3},$$
$\text{when } \sum\limits_{i=1}^3 x_i=n \ \text{and} \ 0<p_i<1 \ \forall \ i\in \ \{1,2,3\}$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .