2) In the absence of other information I'm assuming these probabilities are associated with rolling a fair die. So dataset=1, 2, 3, 4, 5, 6.
Therefore P(X≤4)=4/6=⅔; P(X≥5)=2/6=⅓; P(X=3)=⅙.
MGF is MX(t)=∑etXP(X)=⅙∑etX, because P(X)=⅙ for all X between 1 and 6 (rectangular or uniform distribution).
MX(t) = ⅙(et+e2t+e3t+e4t+e5t+e6t). From this if we put X=3 and t=3 we get: MX=3(t=3)=⅙e9 (not sure if this what is meant in your question).
2b) p(-2)=0.1, p(-1)=k, p(0)=0.2, p(1)=2k, p(2)=0.3, p(3)=k.
If the given values of X and p(X=x) represent the entire population of discrete probabilities, then ∑p(x)=1.
Therefore, 0.1+k+0.2+2k+0.3+k=1, 0.6+4k=1, 4k=0.4, k=0.1.
2b1) If MX(t)=∑etXp(X) for X=-2, -1, 0, 1, 2, 3.
So MX(t)=∑etXp(X)=0.1e-2t+ke-t+0.2+2ket+0.3e2t+ke3t, and, substituting for k:
MX(t)=0.1e-2t+0.1e-t+0.2+0.2et+0.3e2t+0.1e3t.
To find the mean μ=E(X), we need to differentiate MX(t) wrt t to get the first moment:
-0.2e-2t-0.1e-t+0.2et+0.6e2t+0.3e3t, then set t=0: -0.2-0.1+0.2+0.6+0.3=0.8, so mean, μ=0.8.
Variance, σ2, is calculated from the second moment. Differentiate again wrt t:
0.4e-2t+0.1e-t+0.2et+1.2e2t+0.9e3t then set t=0: 0.4+0.1+0.2+1.2+0.9=2.8.
Variance is 2.8.
2b2) To construct the CDF we need to accumulate probabilities:
X: |
-2 |
-1 |
0 |
1 |
2 |
3 |
p(X): |
0.1 |
0.1 |
0.2 |
0.2 |
0.3 |
0.1 |
CDF: |
0.1 |
0.2 |
0.4 |
0.6 |
0.9 |
1.0 |
The CDF graph is represented by the points A to E.
More to follow in due course...