Let X1 and X2 have a trinomial distribution. Differentiate the moment-generating function to
show that their covariance is -np1p2
Solution
The moment generating function is (p1e^t1` + p2e^t2 + 1 - p1 - p2)^n
Using the derivative with repect to t1 and t2 gives us E[x1x2]
Then, var(x1,x2) = E(x1x2) - E(x1)E(x2)
We know that E(x1) = np1 and E(x2) = np2. We can also see this from the first derivative of the
moment generating function.
The derivative with respect to t1 is np1e^t1(p1e^t1` + p2e^t2 + 1 - p1 - p2)^(n-1)
(Note that this gives us E(x1) = np1; calculating for x2 similiarly gives us E(x2) = np2)
Then, the derivative with respect to t2 is
np1e^t1(n-1)p2e^t2(p1e^t1` + p2e^t2 + 1 - p1 - p2)^(n-2)
Evaluating this at t1 = t2 = ...=tk-1 = 0, we getnp1(n-1)p2(1)^n-2=
np1(n-1)p2
Then the covariance isnp1(n-1)p2 - np1(np2) =
p1p2(n^2-n)-n^2p1p2 =
p1p2(n^2-n-n^2) =
-np1p2.
Let X1 and X2 have a trinomial distribution. Differentiate the momen.pdf
1. Let X1 and X2 have a trinomial distribution. Differentiate the moment-generating function to
show that their covariance is -np1p2
Solution
The moment generating function is (p1e^t1` + p2e^t2 + 1 - p1 - p2)^n
Using the derivative with repect to t1 and t2 gives us E[x1x2]
Then, var(x1,x2) = E(x1x2) - E(x1)E(x2)
We know that E(x1) = np1 and E(x2) = np2. We can also see this from the first derivative of the
moment generating function.
The derivative with respect to t1 is np1e^t1(p1e^t1` + p2e^t2 + 1 - p1 - p2)^(n-1)
(Note that this gives us E(x1) = np1; calculating for x2 similiarly gives us E(x2) = np2)
Then, the derivative with respect to t2 is
np1e^t1(n-1)p2e^t2(p1e^t1` + p2e^t2 + 1 - p1 - p2)^(n-2)
Evaluating this at t1 = t2 = ...=tk-1 = 0, we getnp1(n-1)p2(1)^n-2=
np1(n-1)p2
Then the covariance isnp1(n-1)p2 - np1(np2) =
p1p2(n^2-n)-n^2p1p2 =
![Let X1 and X2 have a trinomial distribution. Differentiate the moment-generating function to
show that their covariance is -np1p2
Solution
The moment generating function is (p1e^t1` + p2e^t2 + 1 - p1 - p2)^n
Using the derivative with repect to t1 and t2 gives us E[x1x2]
Then, var(x1,x2) = E(x1x2) - E(x1)E(x2)
We know that E(x1) = np1 and E(x2) = np2. We can also see this from the first derivative of the
moment generating function.
The derivative with respect to t1 is np1e^t1(p1e^t1` + p2e^t2 + 1 - p1 - p2)^(n-1)
(Note that this gives us E(x1) = np1; calculating for x2 similiarly gives us E(x2) = np2)
Then, the derivative with respect to t2 is
np1e^t1(n-1)p2e^t2(p1e^t1` + p2e^t2 + 1 - p1 - p2)^(n-2)
Evaluating this at t1 = t2 = ...=tk-1 = 0, we getnp1(n-1)p2(1)^n-2=
np1(n-1)p2
Then the covariance isnp1(n-1)p2 - np1(np2) =
p1p2(n^2-n)-n^2p1p2 =](https://image.slidesharecdn.com/letx1andx2haveatrinomialdistribution-230324034821-92316a8c/85/Let-X1-and-X2-have-a-trinomial-distribution-Differentiate-the-momen-pdf-1-320.jpg)
