GATE CSE 2017 | Set 2 | Question: 31

Question

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

Comments

Standard Resource for PGF: https://www.stat.auckland.ac.nz/~fewster/325/notes/ch4.pdf

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From the definition of the polynomial function $g(z) = \sum_{j=0}^{N}P(X = j) z^j$, the co-efficient of $z^k$ gives us the probability of random variable $X$ takes value $k$. They have given that for a certain random varible $Y$, the correpondin polynomial function is

$$g(z) = ((1-\beta) + \beta z)^N$$

Using binomial theorem:

\[((1-\beta) + \beta z)^N = \sum_{k=0}^{N}\binom{N}{k}\beta^k(1-\beta)^{n-k}z^k\]

Since the coefficient of $z^k$, gives us the probability of $X$ takes value $k$, Then probability mass function of $Y$ is

\[P(Y=k) = \binom{N}{k}\beta^k (1-\beta)^{N-k}, \text{ for } k = 0, 1, \ldots, N \]

This is nothing but the probability mass function of random variable that follows binomial distribution with parameters $(N, \beta)$. Since the expected value of binomial is product of its parameters, Then

\[E[Y] = N \beta\]

The answer is option B. We don't need any thing about probability generating functions, all we have to recognize the coefficient of $z^k$

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Nice Answer @rupesh17 Thans

Answer 1

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)

 Detailed Video Solution by GO Classes

 

Comments

How does derivative gives us mean?

How do we get this result?

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Any book which has questions dealing with calculus and random variables will have questions like these...if you have calculus intuition you will quickly see that differentiating the polynomial function gives you something close to the expression of the expectation/mean.

$g_{x}(z) = p_{0} + p_{1}z + p_{2}z^{2} + ...$ 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.

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briliant…
u ll do something ossm in GATE 2021

Answer 2

B)

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

Answer 3

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

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Comments

Nicely Explained

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Easy to understand :)

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very good explanation but very difficult that it will click in normal exam like scenario

Answer 4

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)

$= E(x) = N\beta(1 – \beta + \beta)$

$= E(x) = N\beta$

$\therefore Answer$ $is$ $B$

 

Comments

In the question E[Y]  is to be answered but here it’s mentioned E[X]. Is this a mistake or am I missing something? @Soumya Saurav

Answer 5

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 6

This question is one of the toughest questions asked in gate from probability – Sachin sir

Here is a video solution

https://youtu.be/5XsTMkXpdfA