we will consider a random variable whose pdf is the form of f(x) = Cx™(1-x) where x E (0, 1) and m is a non-negative integer. (a) Consider the special situation of m = 1. Find the value of C such that the pdf is a proper pdf. (b) Argue that for a general m, the pdf must be f (x) = (m + 1)(m + 2)x™(1 – x) for x E (0, 1).

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we will consider a random variable whose pdf is the form of f(x) = Cx"(1-x)
where x E (0, 1) and m is a non-negative integer.
(a) Consider the special situation of m =
1. Find the value of C such that the pdf is a proper
pdf.
(b) Argue that for a general m, the pdf must be
f(x) = (m + 1)(m + 2)x"(1 – x) for x E (0,1).
(c) Show that the cumulative distribution function (CDF) of X is
xm+1
xm+2
F(x) = (m + 1)(m + 2)
т+1
т +2
(d) Continue from part (c). Evaluate the CDF at -1, 0.5, and 2 i.e., find F(-1), F(0.5), and
F(2). Be careful and think twice about your answers.
(e) In this class, we haven't studied the concept of expectation. For a continuous random
variable X, its expectation is defined as E[X] = o xf(x)dx. Find E[X].
Transcribed Image Text:we will consider a random variable whose pdf is the form of f(x) = Cx"(1-x) where x E (0, 1) and m is a non-negative integer. (a) Consider the special situation of m = 1. Find the value of C such that the pdf is a proper pdf. (b) Argue that for a general m, the pdf must be f(x) = (m + 1)(m + 2)x"(1 – x) for x E (0,1). (c) Show that the cumulative distribution function (CDF) of X is xm+1 xm+2 F(x) = (m + 1)(m + 2) т+1 т +2 (d) Continue from part (c). Evaluate the CDF at -1, 0.5, and 2 i.e., find F(-1), F(0.5), and F(2). Be careful and think twice about your answers. (e) In this class, we haven't studied the concept of expectation. For a continuous random variable X, its expectation is defined as E[X] = o xf(x)dx. Find E[X].
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