The lifetime (in months) of a battery is modeled by a random variable X that has pdf

Question

\displaystyle \displaystyle f_\theta (x)=K\theta ^ x \mathbf{1}(x>0)\qquad \text {where }\, K=\ln (1/\theta )
for an unknown parameter \theta \in (0,1). (Here \mathbf{1}(x>0) is the indicator variable that takes value 1 when its argument is true, i.e. when x>0.)

Assume that we have n independent observations X_1, \ldots , X_ n of the lifetime of n batteries of the same type. We want to use these observations to estimate \theta \in (0,1).

Compute the expected value \mathbb E[X_ i] and the variance \textsf{Var}[X_ i] of X_ i.

(Enter your answer in terms of \theta only. )

Hint: Note that the given pdf is equivalent to the pdf of a common distribution except reparametrized with a different parameter.

Answer 1

The given pdf is in the form of a geometric distribution, with parameter p = \theta. The probability mass function of the geometric distribution is given by:

P(X = x) = pq^(x-1)

Where q = 1 - p.

In this case, p = \theta and q = 1 - \theta.

The expected value of a geometric distribution is given by:

E(X) = 1/p = 1/\theta

So, in this case, the expected value of X_i is:

E(X_i) = 1/\theta

The variance of a geometric distribution is given by:

Var(X) = q/p^2 = (1 - \theta)/\theta^2

So, in this case, the variance of X_i is:

Var(X_i) = (1 - \theta)/\theta^2

Answer 2

are you sure

Answer 3

I apologize for the mistake in my previous response.

The given pdf is a discrete exponential distribution, with parameter λ = -ln(θ). The probability mass function of the discrete exponential distribution is given by:

P(X = x) = λθ^x for x > 0

The expected value of a discrete exponential distribution is given by:

E(X) = 1/λ = -1/ln(θ)

So, in this case, the expected value of X_i is:

E(X_i) = -1/ln(θ)

The variance of a discrete exponential distribution is given by:

Var(X) = (1 - θ)/(λ^2 * (1 - θ)^2)

Using λ = -ln(θ), we can rewrite the variance as:

Var(X) = (1 - θ)/(ln^2(θ) * (1 - θ)^2)

So, in this case, the variance of X_i is:

Var(X_i) = (1 - θ)/(ln^2(θ) * (1 - θ)^2)

Answer 4

are you sure

Answer 5

I apologize for the confusion. Let's re-evaluate the problem.

The probability density function (pdf) given is:

f_θ(x) = Kθ^x * 1(x > 0)

where K = ln(1/θ)

To compute the expected value E[X_i], we integrate x times the pdf with respect to x over the range 0 to infinity:

E[X_i] = ∫[0,∞] x * Kθ^x dx

To find the variance Var[X_i], we need to compute E[X_i^2] first:

E[X_i^2] = ∫[0,∞] x^2 * Kθ^x dx

Then we can use the formula:

Var[X_i] = E[X_i^2] - (E[X_i])^2

Let's proceed with the calculations.