expanding above :M(t) =(1/36)*e-6t +(25/36)e4t +(10/36)e-t
therefore
P(X=-6)=1/36
P(X=-1) =10/36
P(X=4)=25/36
P(X>-3|X<2) =P(-3<X<2)/P(X<2) =(10/36)/(11/36)=10/11 =0.91
3. Let X be a random variable and denote by Mx(t) its MGF. Prove that, for any t > 0, we have P[X >Mx(t)e
2.4.10 A random variable Xhas its mgf given by Mx(t) e (5 - 4e')1 for t< 223. Evaluate P(4 or 5). Hint: What is the mgf of a geometric random vari- able?
EXERCISES the mgf of the random variable X, find P(x < 5.23). d so that P(e < X < d) = 0.95 3.3.1. If (1-2t)-6, t<½, is V3. .3.2. If X is χ2 (5), determine the constants c and and P(X < c) = 0.025.
The mgf of a random variable X has the following form: e-8t et 5 Mx(t) = 0.64 . Find ElYX). Answer:-0.2
Let X be a random variable with MGF MX(t)=13e−t+16+12e2t. What is P(X≤1)?
3. Let X be a random variable and denote by Mx(t) its MGF. Prove that, for any t > 0, we have
3. Let X be a random variable and denote by Mx(t) its MGF. Prove that, for any t > 0, we have
The moment generating function (MGF) for a random variable X is: Mx (t) = E[e'X]. Onc useful property of moment generating functions is that they make it relatively casy to compute weighted sums of independent random variables: Z=aX+BY M26) - Mx(at)My (Bt). (A) Derive the MGF for a Poisson random variable X with parameter 1. (B) Let X be a Poisson random variable with parameter 1, as above, and let y be a Poisson random variable with parameter y. X...
(3 marks) The moment generating function of a random variable X is given by MX(t) = 24 20 < - In 0.6. Find the mean and standard deviation of X using its moment generating function.
A random variable X has the following mgf
et
M(t)=1−t, t<1.
(a) Find the value of ∞ (−1)k E(Xk).
(b) Find the value of E(2−X).
(c) Find the value of Var(2−X).
(d) Find the probability P (X > 4).
10. A random variable X has the following mgf М() t 1 1 t (a) Find the value of 1E(Xk) (b) Find the value of E(2X). (c) Find the value of Var(2-X) k 0 k! (d) Find the probability P(X >...
If Mx(t) e-51-e), find Var(X)