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49 votes
49 votes

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
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7 Answers

Best answer
55 votes
55 votes

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)

edited by
5 votes
5 votes

B)

If you expand gy(z) you get binomial distribution and mean of binomial distribution is N*p.

3 votes
3 votes

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)

$= E(x) = N\beta(1 – \beta + \beta)$

$= E(x) = N\beta$

$\therefore Answer$ $is$ $B$

 

1 votes
1 votes

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...

Answer:

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