*6. In lecture it is claimed that since "-1S² 2→ o² and „", → 1, their product converges inprobability to o²; in other words, that the product of a convergent sequence of real numbersand a sequence of random variables with a limit in probability converges in probability to theproduct of the limits.Suppose that {Yn} is a sequence of random variables that converges in probability to Y:lim P (|Y, – Y| > e) = 0Ve > 0n-00Let {an} be a sequence of real numbers that converges to a. That is, for any e > 0, there issome n* such thatn > n*|an – al < eShow that a, Yn 2» aY. (Hint: show that an » a and use the property (without proof)that convergence in probability is preserved under multiplication.)

Given: Yn is a sequence of random variables that converges in probability to Y. So, limn→∞PYn-Y>ε=0 ∀ε>0. an is a sequence of real numbers that converge to a. So, for any ε>0, there is some n* such that n>n*⇒an-a<ε. To show: anYn→paYFirst, we will try to relate a sequence of random variables to a sequence of real numbers. In any probability model, there is a sample space S and a probability measure P. Now, a random variable is a mapping that assigns a real number to any of the possible outcomes. That is, a random variable is a mapping that assigns a real number to each of the elements of the sample space S. When there is a sequence of random variables, X1,X2,...,Xn, there is also an underlying sample space. In particular, each Xn is a function from S to ℝ. So, a sequence of random variables is in fact a sequence of functions, Xn:S→ℝ. Now, let each Xn be a constant function such that Xns=an ∀s∈S Then, each random variable Xn in the sequence of random variables Xn is equivalent to (or in other words, has a direct correlation to) a term an in the sequence of real numbers an.Now, coming back to the problem at hand. Let Xn be a sequence of random variables such that Xns=an ∀s∈S or equivalently, Xn=an. It is given that an converges to a. So, for any ε>0, there is some n* such that n>n*⇒an-a<ε This implies that limn→∞an=a⇒limn→∞an-a=0 Let X be a random variable such that Xs=a ∀s∈S or equivalently, X=a. Put Xn=an and X=a in limn→∞an-a=0 to get, limn→∞Xn-X=0, which implies, limn→∞PXn-X>ε=0 ∀ε>0 This implies that Xn→pX or equivalently, an→paNow, we know that convergence in probability is preserved under multiplication. That is, if Xn→pX and Yn→pY then XnYn→pXY. It is given that Yn→pY and as shown previously, an→pa. Then, we have, anYn→paY. This is the required proof.Answer: It has been shown that anYn→paY.