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Official PYQ · 2017

GATE CS 2017 Set 2 Q31 — Probability: Random Variables and Distributions

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For any discrete random variable XX, with probability mass function P(X=j)=pj,pj≥0,j∈{0,…,N}P(X=j)=p_j, p_j \geq 0, j \in \{0, \dots , N \}, and Σj=0N pj=1\Sigma_{j=0}^N \: p_j =1, define the polynomial function gx(z)=Σj=0N pj zjg_x(z) = \Sigma_{j=0}^N \: p_j \: z^j. For a certain discrete random variable YY, there exists a scalar β∈[0,1]\beta \in [0,1] such that gy(z)=(1−β+βz)Ng_y(z) =(1- \beta+\beta z)^N. The expectation of YY is

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Answer: B

The given polynomial function gX(z)=∑j=0Npjzjg_X(z) = \sum_{j=0}^N p_j z^j is the Probability Generating Function (PGF) of the discrete random variable XX.

For the random variable YY, the PGF is given as: gY(z)=(1−β+βz)Ng_Y(z) = (1 - \beta + \beta z)^N

Method 1: Using the derivative of the PGF

The expectation (mean) of a discrete random variable YY can be computed by evaluating the first derivative of its PGF at z=1z = 1: E[Y]=gY′(1)E[Y] = g'_Y(1)

Differentiating gY(z)g_Y(z) with respect to zz: gY′(z)=ddz[(1−β+βz)N]g'_Y(z) = \frac{d}{dz} \left[ (1 - \beta + \beta z)^N \right] gY′(z)=N(1−β+βz)N−1⋅βg'_Y(z) = N(1 - \beta + \beta z)^{N-1} \cdot \beta

Evaluating this derivative at z=1z = 1: E[Y]=gY′(1)=N(1−β+β(1))N−1⋅βE[Y] = g'_Y(1) = N(1 - \beta + \beta(1))^{N-1} \cdot \beta E[Y]=N(1)N−1⋅β=NβE[Y] = N(1)^{N-1} \cdot \beta = N\beta

Method 2: Recognizing the distribution

The PGF of a Binomial random variable with parameters NN (number of trials) and pp (probability of success) is: G(z)=(q+pz)N=(1−p+pz)NG(z) = (q + pz)^N = (1 - p + pz)^N

Comparing this with gY(z)=(1−β+βz)Ng_Y(z) = (1 - \beta + \beta z)^N, we can see that YY is a Binomial random variable with parameters NN and p=βp = \beta.

The expectation of a Binomial random variable is: E[Y]=Np=NβE[Y] = Np = N\beta

Thus, the correct option is B.

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