Exercise: Convergence in probability
a) Suppose that Xn is an exponential random variable with parameter λ=n. Does the sequence {Xn} converge in probability?
b) Suppose that Xn is an exponential random variable with parameter λ=1/n. Does the sequence {Xn} converge in probability?
c) Suppose that the random variables in the sequence {Xn} are independent, and that the sequence converges to some number a, in probability. Let {Yn} be another sequence of random variables that are dependent, but where each Yn has the same distribution (CDF) as Xn. Is it necessarily true that the sequence {Yn} converges to a in probability?
a) Yes
b) No
c) Yes
a) Well, my odds of giving a good answer to this are definitely converging to zero. Anyways, let's see. If Xn is an exponential random variable with parameter λ=n, it means that as n goes to infinity, the rate of decay of Xn also increases. So, does this sequence converge in probability? Umm... I'm not great at math, but intuitively, as λ increases, Xn tends to zero. So, I would guess that the sequence {Xn} does converge in probability, but it might take a while. Just like trying to find a parking spot in a crowded lot.
b) Now we have Xn as an exponential random variable with parameter λ=1/n. Hmmm... So, as n increases, the rate of decay decreases. It's like trying to sprint with a snail as your coach. Well, in this case, the sequence {Xn} probably doesn't converge in probability. It's more like a never-ending rollercoaster ride, going up and down without any stable destination.
c) Ah, now we have some twist in the tale. The random variables in the sequence {Xn} converge to a, in probability. And then we have {Yn}, a sequence of dependent random variables with the same distribution as Xn. Will {Yn} also converge to a in probability? Hmm, well, dependency can be a real deal-breaker. Just like when you depend on your alarm clock to wake you up, but it decides to take a vacation. So, in this case, the convergence of {Yn} to a might depend on the degree of dependency between the variables. It's as uncertain as predicting the weather when the meteorologist is on vacation.
a) To determine if the sequence {Xn} converges in probability, we need to check if it satisfies the definition of convergence in probability.
Recall that a sequence of random variables {Xn} converges in probability to a constant c if, for any positive value ε, the probability that Xn deviates from c by more than ε tends to zero as n approaches infinity.
In this case, Xn is an exponential random variable with parameter λ=n. The probability density function of an exponential random variable is given by f(x) = λe^(-λx), where λ is the rate parameter.
To find the probability that Xn deviates from a constant c by more than ε, we can calculate the integral of the probability density function over the interval (c-ε, c+ε).
P(|Xn-c| > ε) = ∫_(c-ε)^(c+ε) λe^(-λx) dx
= ∫_(c-ε)^(c+ε) ne^(-nx) dx
= [-e^(-nx)]_(c-ε)^(c+ε)
= e^(-n(c-ε)) - e^(-n(c+ε))
As n approaches infinity, the exponential terms tend to zero, so for any fixed ε, P(|Xn-c| > ε) tends to zero. Therefore, the sequence {Xn} converges in probability to the constant c=0.
b) Similarly, for the sequence {Xn} where Xn is an exponential random variable with parameter λ=1/n, we can follow the same steps as in part (a).
P(|Xn-c| > ε) = ∫_(c-ε)^(c+ε) (1/n)e^(-x/n) dx
= [-ne^(-x/n)]_(c-ε)^(c+ε)
= ne^(-n(c-ε)/n) - ne^(-n(c+ε)/n)
= e^(-n(c-ε)) - e^(-n(c+ε))
Again, as n approaches infinity, the exponential terms tend to zero, and for any fixed ε, P(|Xn-c| > ε) tends to zero. Therefore, the sequence {Xn} also converges in probability to the constant c=0.
c) If the sequence {Xn} converges to some number a, in probability, and {Yn} is another sequence of random variables that are dependent, but each Yn has the same distribution (CDF) as Xn, it is not necessarily true that the sequence {Yn} converges to a in probability.
Convergence in probability depends on the joint distribution of the sequence of random variables. While the individual random variables {Xn} and {Yn} may have the same distribution, the dependence between the random variables in the sequence {Yn} can affect convergence behavior.
Therefore, it is not guaranteed that the sequence {Yn} converges to the same constant a in probability as the sequence {Xn}.
a) To determine if the sequence {Xn} converges in probability, we need to check if the probability that Xn is within a given distance ε of some fixed number a approaches 1 as n approaches infinity.
In this case, Xn is an exponential random variable with parameter λ=n. The probability density function (PDF) of an exponential random variable with parameter λ is given by f(x) = λ * exp(-λx) for x >= 0.
To find the probability that Xn is within ε of a, we need to integrate the PDF over the interval (a-ε, a+ε) and see if this approaches 1 as n approaches infinity.
Let's calculate this integral for Xn:
P(|Xn - a| < ε) = ∫[a-ε, a+ε] λ * exp(-λx) dx
The integral can be evaluated as follows:
P(|Xn - a| < ε) = [-exp(-λx)] evaluated from a-ε to a+ε
P(|Xn - a| < ε) = -exp(-λ(a+ε)) + exp(-λ(a-ε))
Now, since λ=n, we can substitute n for λ:
P(|Xn - a| < ε) = -exp(-n(a+ε)) + exp(-n(a-ε))
To check if the sequence {Xn} converges in probability, we need to determine if this probability approaches 1 as n approaches infinity.
b) Following a similar approach as in part a), we have Xn as an exponential random variable with parameter λ=1/n. The PDF of Xn is given by f(x) = (1/n) * exp(-(1/n)x) for x >= 0.
Let's calculate the probability that Xn is within ε of a:
P(|Xn - a| < ε) = ∫[a-ε, a+ε] (1/n) * exp(-(1/n)x) dx
The integral can be evaluated as follows:
P(|Xn - a| < ε) = [-exp(-(1/n)x)] evaluated from a-ε to a+ε
P(|Xn - a| < ε) = -exp(-(1/n)(a+ε)) + exp(-(1/n)(a-ε))
To check if the sequence {Xn} converges in probability, we need to determine if this probability approaches 1 as n approaches infinity.
c) In this case, we have a sequence {Xn} that converges to some number a in probability. Let {Yn} be another sequence of random variables that are dependent, with each Yn having the same distribution (CDF) as Xn.
To determine if the sequence {Yn} necessarily converges to a in probability, we need to consider the dependence between the variables.
The fact that the variables are independent in the sequence {Xn} means that the convergence in probability of {Xn} is solely determined by the properties of each individual variable.
However, when we move to the dependent sequence {Yn}, the dependence between the variables can impact the convergence behavior. The dependencies can introduce additional fluctuations and variations in the sequence, potentially affecting the convergence properties.
As a result, it is not necessarily true that the sequence {Yn} will converge to a in probability, even if each Yn has the same distribution as Xn. The dependence between the variables can introduce additional complexities that may prevent convergence or lead to a different limit. To determine the convergence behavior of {Yn}, the specific dependence structure needs to be considered.