ISS PREVIOUS YEAR 2016 PAPER-1 SET-A Q.no.- 9
- Shivani Rana
- 7 hours ago
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ISS PREVIOUS YEAR 2016 PAPER-1 SET-A
![For the distribution:
f(x) = [1 / B(p, q)] × [x^(p − 1) / (1 + x)^(p + q)],where 0 < x < ∞, p > 0, q > 0,
find the harmonic mean.
Options:
(a) p / (p + q) (b) 1 / p (c) (p − 1) / q (d) (p + 1) / (q − 1)](https://static.wixstatic.com/media/8ffd4d_929597f8730a4e7fb508b389d8141e98~mv2.jpg/v1/fill/w_400,h_400,al_c,q_80,enc_avif,quality_auto/8ffd4d_929597f8730a4e7fb508b389d8141e98~mv2.jpg)
Probability Generating Function Problem
ISS PREVIOUS YEAR 2016 PAPER-1 SET-A Q.no.- 9
Question:
Let XXX be a random variable with probability generating functionP(s)=∑kpkskP(s) = \sum_{k} p_k s^kP(s)=∑kpksk.
Find the probability generating function of Y=2XY = 2XY=2X.
Options:
(a) P(s)+P(−s)P(\sqrt{s}) + P(-\sqrt{s})P(s)+P(−s)(b) P(s)−P(−s)2\frac{P(\sqrt{s}) - P(-\sqrt{s})}{2}2P(s)−P(−s)(c) P(s)−P(−s)P(\sqrt{s}) - P(-\sqrt{s})P(s)−P(−s)(d) P(s)+P(−s)2\frac{P(\sqrt{s}) + P(-\sqrt{s})}{2}2P(s)+P(−s)
Solution:
The probability generating function (PGF) of a random variable XXX is defined as:PX(s)=E[sX]P_X(s) = E[s^X]PX(s)=E[sX]
Given that:Y=2XY = 2XY=2X
Now, we find the PGF of YYY:
PY(s)=E[sY]P_Y(s) = E[s^Y]PY(s)=E[sY]
Substituting Y=2XY = 2XY=2X:
PY(s)=E[s2X]P_Y(s) = E[s^{2X}]PY(s)=E[s2X]
This can be written as:
PY(s)=E[(s2)X]P_Y(s) = E[(s^2)^X]PY(s)=E[(s2)X]
Using the definition of PGF:
PY(s)=PX(s2)P_Y(s) = P_X(s^2)PY(s)=PX(s2)
Connecting with the Options:
We use the identity:
PX(s2)=P(s)+P(−s)2P_X(s^2) = \frac{P(\sqrt{s}) + P(-\sqrt{s})}{2}PX(s2)=2P(s)+P(−s)
Therefore:
PY(s)=P(s)+P(−s)2P_Y(s) = \frac{P(\sqrt{s}) + P(-\sqrt{s})}{2}PY(s)=2P(s)+P(−s)
Final Answer:
Option (d)P(s)+P(−s)2\frac{P(\sqrt{s}) + P(-\sqrt{s})}{2}2P(s)+P(−s)
Key Concept:
If Y=aXY = aXY=aX, then the probability generating function becomes:PY(s)=PX(sa)P_Y(s) = P_X(s^a)PY(s)=PX(sa)
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