For any discrete random variable \(X\), with probability mass function…
2017
For any discrete random variable \(X\), with probability mass function
\(P(X=j)=p_j, p_j \geq 0, j \in \{0, \dots , N \}\), and \(\Sigma_{j=0}^N \: p_j =1\), define the polynomial function \(g_x(z) = \Sigma_{j=0}^N \: p_j \: z^j\). For a certain discrete random variable \(Y\), there exists a scalar \(\beta \in [0,1]\) such that \(g_y(z) =(1- \beta+\beta z)^N\). The expectation of \(Y\) is
- A.
\(N \beta(1-\beta)\) - B.
\(N \beta\) - C.
\(N (1-\beta)\) - D.
Not expressible in terms of
\(N\)and\(\beta\)alone
Attempted by 24 students.
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Correct answer: B
Solution:
Key idea: The expectation of a discrete random variable can be found from its probability-generating function g_y(z)=E[z^Y] by differentiation: E[Y]=g_y'(1).
Differentiate the given generating function g_y(z) = (1 - β + β z)^N with respect to z.
Compute g_y'(z) = N β (1 - β + β z)^{N-1}.
Evaluate at z = 1 to get E[Y] = g_y'(1) = N β.
Alternatively, expand the polynomial: (1 - β + β z)^N = Σ_{j=0}^N C(N,j)(1-β)^{N-j}β^j z^j, so Y has the Binomial(N, β) distribution and its mean is Nβ.
Therefore, the expectation of Y is Nβ.