Let X1, X2, …, Xn be a sequence of discrete random variables...

Let $X_1, X_2, \dots, X_n$ be a sequence of discrete random variables with PDFs $f_1(x), f_2(x), \dots, f_n(x)$ with corresponding support sets $D_i = \text{supp}(f_i)$ for $i = 1, 2, \dots, n$. Pick a sequence of real numbers $p_1, p_2, \dots, p_n$ that satisfies the following two conditions: i) $p_i \ge 0$ for each $i = 1, 2, \dots, n$ ii) $\sum_{i=1}^n p_i = 1$. Show that the function $f(x) = \sum_{i=1}^n p_i f_i(x)$ is a PDF.

Transcript

00:01 Hello there, welcome to number one we are going to cover this question relating to bounded random variable we have the sequence x and converges to a random variable xxxifounded and we have to prove that it also converges in mean to x right, so how do we do? so we're gonna write down the definition of expectation expectation here is equal to the what is equal to the iteration right? of a measurable set omega of the absolute values of x n minus x right the omega so you can consider a small omega here the omega right so now you can separate the set omega into two subset one set is when x n minus x bigger than epsilon and another set is x and minus x is let them or equal to epsilon right okay so what do we do next? all right, so on the left hand side, on the first time, i'm going to apply this property, this assumption, okay, and this assumption and this assumption, right? so you can do the analysis, right? so x -n minus x less than absolute value of x -n plus x, and x -n is bounded by mu, and x -a also bounded by mu, so we have two mu, right, times the iteration of this.

01:34 So it's going to be the probability, that x n minus x bigger than epsilon right on the other hand the second term we're going to deal with this one right so how do we do this right so x n minus x is less than epsilon right so therefore we have epsilon here so it's led than epsilon here and then the measure of this set and the measure of this set is x n minus x let than epsilon right and the total probability equal to 1 right so the probability here equal to 1 the whole probability of omega equal to 1 so this is less than 1 so it's strictly less than 2m time probability that x n minus x bigger than epsilon plus epsilon right and now now because xn converges to x in probability so when you because of that then you can you can choose in is big enough, right? or sufficiently large, sufficiently large.

02:46 Then you have this term less than or equal to 2 mu time epsilon, right? and then plus epsilon.

02:53 So you can factor out as the epsilon, you have 2m plus 1...

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