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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

1. $N \beta(1-\beta)$
2. $N \beta$
3. $N (1-\beta)$
4. Not expressible in terms of $N$ and $\beta$ alone

@rupesh17

How did you take P(Y = 1) = $\beta$ ? And why does Y have only two values (0,1) ?
edited by

see polynomial function g(z) is defined like that….

$g_{x}(z)=\sum_{0}^{N}p_{j}z^{^{j}}$ for  P(X=j)=$p_{j}$, j∈{0,…,N}

Meaning-

suppose  variable x  has probability p0,p1,p2..at x=0 x=1 and x=2 respectively then

$g_{x}(z)=p_{0}+p_{1}z+p_{2}z^{2}+...$  by definition of given function….

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now we are given that—

$g{_{y}}$(z)=$(1−β+βz)^{^{N}}$ and we assume N=1

then

$g{_{y}}$(z)=1−β+βz

so

coefficient of $z^{^{0}}$ gives probability at y=0 i.e p0 and

coefficient of z gives probability at y=1 i.e p1 .

Expectation of Y is β at N=1 ..only option B matching…..

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We can verify by taking any value eg. N=2

rupesh17

Very nice. Thanks for the explanation

This is the toughest question of 2017 GATE exam but you can easily understand the answer like this….…..

B option only mean of binomial distribution is np

Can you please provide explanation for this?
Can some one explain this in some easy way?