Let Y denote a random variable that has a Poisson
distribution with mean λ = 6. (Round your answers to three
decimal places.)
(a) Find P(Y = 9).
(b) Find P(Y ≥ 9).
(c) Find P(Y < 9).
(d) Find P(Y ≥
9|Y ≥ 6).
Let Y denote a random variable that has a Poisson distribution
with mean λ = 3. (Round your answers to three decimal places.)
(a) Find P(Y = 6)
(b) Find P(Y ≥ 6)
(c) Find P(Y < 6)
(d) Find P(Y ≥ 6|Y ≥ 3).
Suppose X is an exponential random variable with mean 5 and Y is
an exponential random variable with mean 10. X and Y are
independent. Determine the coefficient of variation of X + Y
X and Y are independent Exponential random variables with
mean=4, λ = 1/2.
1) Find the joint CDF of the random variables X, Y
and Find the probability that 4X > Y .
2) Find the expected value of X^3 + X*Y .
1. Assume Y is an exponential random variable with rate
parameter λ=2.
(1) Generate 1000 samples from this exponential distribution
using inverse transform method
(2) Compare the histogram of your samples with the true density
of Y.
Please provide MATLAB programming code below with graphs if
possible. Thank you!
The exponential distribution with rate λ has mean μ = 1/λ. Thus
the method of moments estimator of λ is 1/X. Use the following
steps to verify that X is unbiased, but 1/X is biased.
a) Generate 10000 samples of size n = 5 from the standard
exponential distribution (i.e. λ = 1) using rexp(50000) and
arranging the 50000 random numbers in a matrix with 5 rows.
b) Use the apply() function to compute the 10000 sample means
and store...
Let p(y) denote the probability function associated with a
Poisson random variable with mean λ. Show that the ratio of
successive probabilities satisfies (p(y)/p(y-1)) = ( λ /y), for
y=1,2,...
For a random sample of sizenfrom an Exponential distribution
with rate parameter λ (so that the density is fY(y) =λe−λy), derive
the maximum likelihood estimator, the methods of moments estimator,
and the Bayes estimator (that is, the posterior mean) using a prior
proportional to λe−λ, for λ >0. (Hint: the posterior
distribution will be a Gamma.)
This is the code for the LLN question iid random variable from
exponential distribution with mean beta. Beta is fixed 1 for
illustration.
# Testing LLN
plot.ybar <- function(n, m = 1e5, beta = 1) {
Ybars <- rep(0,m)
for(i in 1:m) {
Y.sample <- rexp(n,1/beta)
Ybars[i] <- mean(Y.sample)
}
p <- hist(Ybars, xlim=c(0,4), freq=F,main = paste("Distribution
of Ybar when n=",n, sep=""))
}
par(mfrow = c(2,2))
for (n in c(2, 10, 50, 200)) {
plot.ybar(n)
}
Using similar code...
Given the cumulative distribution of an exponential random
variable find:
The probability density function
Show that it is a valid probability function
The moment generating function
The Expected mean
The variance