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

- $N \beta(1-\beta)$
- $N \beta$
- $N (1-\beta)$
- Not expressible in terms of $N$ and $\beta$ alone

### 9 Comments

Here is one way to solve the question –

https://gateoverflow.in/?qa=blob&qa_blobid=14447251432178239340

(Not putting this as an answer since I have not solved completely and only written steps)

Pascua 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

## 5 Answers

Notice that the derivative of $\large g{_x}(z)$ evaluated at $z=1$ gives expectation $E(X)$

$g'_x(z)|_{z=1}= \Sigma_{j=1}^N\,j\, p_j\, z^{j-1} |_{z=1} = \Sigma_{j=1}^N\,j\, p_j = \Sigma_{j=0}^N\,j\, p_j = E(X)$

Therefore, take derivative of $\large g{_y}(z)$ with respect to $z,$ and plug in $z=1$

$E(Y) = g'_y(z)|_{z=1}= ((1-\beta + \beta\,z)^N )' |_{z=1} = N\beta(1-\beta + \beta\,z)^{N-1}|_{z=1} = N\beta(1-\beta + \beta)^{N-1} = N\beta$

**So, Answer is option (B)**

### 10 Comments

$g_{x}(z) = p_{0} + p_{1}z + p_{2}z^{2} + ...$ by definition of the polynomial function. Differentiate this to get -:

$g'_{x}(z) = 0 + p_{1} + 2p_{2}z + 3p_{3}z^{2} + ...$

Now recall that the expectation of a discrete random variable is defined as -:

$E[X] = \sum_{x = 0}^{N}xp(x) = 0 + p_{1} + 2p_{2} + 3p_{3} + ...$

So if we could just remove the $z$ terms from our differentiated polynomial function, we would have $E[X]$. To do that, just calculate $g'_{x}(1) = p_{1} + 2p_{2} + ...$

We are given a closed-form expression for $Y$'s polynomial function. Follow the same logic - differentiate it and then apply $z = 1$ to get the answer.

The following solution is based on the idea shared by @Sachin Mittal sir.

Given,

$g_y(z) = (1 – \beta + \beta z)^N$

$= g_y(z) = ((1 – \beta) + \beta z)^N$

Using binomial expansion,

$= g_y(z) = \sum_{k=0}^{N}{\binom{N}{k}(1 – \beta)^{N-k}(\beta z)^k}$

$= g_y(z) = \sum_{k=0}^{N}{\binom{N}{k}(1 – \beta)^{N-k}(\beta)^k(z)^k}$

Let $a = (1 – \beta)$ and $b = \beta$

$= g_y(z) = \sum_{k=0}^{N}{\binom{N}{k}a^{N-k}b^k(z)^k}$

comapring $g_x(z) = \sum_{j=0}^{N}p_jz^j$ to the above $eq^n$ we know that each of the terms in the above summation will give $P(Y = j) = p_j, where$ $0 \leq j \leq N$

and $\because E(x) = \sum_{k=0}^{N}k*p_k$

$= E(x) = \sum_{k=0}^{N}{\binom{N}{k}a^{N-k}b^{k}k}$

$= E(x) = Nb(a + b)^{N-1}$

(https://www.wolframalpha.com/input/?i=sum+ncr%28n%2C+k%29+*+x%5E%28n-k%29*y%5Ek*k%2Ck%3D0+to+n)

$= E(x) = N\beta(1 – \beta + \beta)$

$= E(x) = N\beta$

$\therefore Answer$ $is$ $B$

For a discrete random variable X,

Given gy (z) = (1 - β + βz)N ⇾ it is a binomial distribution like (x+y)n

Expectation (i.e., mean) of a binomial distribution will be np.

The polynomial function ,

given

Mean of Binomial distribution of b(xj,n,p)=

The probability Mass function,

Given:

Hope this helps...