# Thread: trinomial distribution

1. ## trinomial distribution

Let X and Y have a trinomial distribution with parameters n=3, $p_{1}=\frac{1}{6}$ and $p_{2}=\frac{1}{2}$. Find:
a) E(X)
b) E(Y)
c) Var(X)
d) Var(Y)
e) Cov(X,Y)
f) $\rho$

*Note that $\rho=-\sqrt{p_{1}p_{2}/(1-p_{1})(1-p_{2})}$

2. Originally Posted by antman
Let X and Y have a trinomial distribution with parameters n=3, $p_{1}=\frac{1}{6}$ and $p_{2}=\frac{1}{2}$. Find:
a) E(X)
$E(X)=np_1=3\left(\frac{1}{6}\right)=\boxed{\frac{1 }{2}}$

b) E(Y)
Similarly, $E(Y)=np_2=3\left(\frac{1}{2}\right)=\boxed{\frac{3 }{2}}$

c) Var(X)
$\text{Var}(x)=np_1(1-p_1)=3\left(\frac{1}{6}\right)\left(\frac{5}{6}\ri ght)=\boxed{\frac{5}{12}}$

d) Var(Y)
Similarly, $\text{Var}(Y)=np_2(1-p_2)=3\left(\frac{1}{2}\right)\left(\frac{1}{2}\ri ght)=\boxed{\frac{3}{4}}$

e) Cov(X,Y)
I'm not too sure about the Covariance. I'm tempted to say that its $-np_1p_2=-3\left(\frac{1}{6}\right)\left(\frac{1}{2}\right)= \boxed{-\frac{1}{4}}$ (Someone verify this... -_-)

f) $\rho$

*Note that $\rho=-\sqrt{p_{1}p_{2}/(1-p_{1})(1-p_{2})}$
You're given enough information to find $\rho$. Its just plug-n-chug from here...

Does this make sense?

3. Originally Posted by Chris L T521
I'm not too sure about the Covariance. I'm tempted to say that its $-np_1p_2=-3\left(\frac{1}{6}\right)\left(\frac{1}{2}\right)= \boxed{-\frac{1}{4}}$ (Someone verify this... -_-)
For one trial, $\mathbb{E}(X_1Y_1)=\mathbb{P}(X_1=1,Y_1=1)=0$, because both can't happen at the same time.

$\text{cov}(X_1,Y_1)=\mathbb{E}(X_1Y_1)-\mathbb{E}(X_1)\mathbb{E}(Y_1)=-p_1p_2$

Then, since in the trinomial distribution we have n independent trials, $\text{cov}(X,Y)=-n\text{cov}(X_1,Y_1)=-np_1p_2$
( $X=\sum_{i=1}^n X_i$ and $Y=\sum_{i=1}^n Y_i$)

,

# let x and y have a trinomial distribution with parameters n 3

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