E(X) & Var(X) of PGFs (Edexcel A Level Further Maths: Further Statistics 1): Revision Note

Exam code: 9FM0

Mark Curtis

Written by: Mark Curtis

Reviewed by: Dan Finlay

Updated on

E(X) of PGFs

How do I find E(X) of a PGF?

  • E(X)=GX'(1)

    • Differentiate the PGF and substitute in t=1

  • This is because 

    • straight G subscript X open parentheses t close parentheses equals straight E open parentheses t to the power of X close parentheses equals sum from blank to blank of space t to the power of x space straight P open parentheses X equals x close parentheses

    • So straight G subscript X apostrophe open parentheses t close parentheses equals stack fraction numerator straight d over denominator straight d t end fraction open square brackets sum from blank to blank of space t to the power of x space straight P open parentheses X equals x close parentheses close square brackets equals sum with blank below and blank on top space x t to the power of x minus 1 end exponent space straight P open parentheses X equals x close parentheses

    • Substituting t=1 gives straight G subscript X apostrophe open parentheses 1 close parentheses equals sum from blank to blank of space x space straight P open parentheses X equals x close parentheses space equals space straight E open parentheses X close parentheses

  • You may need the chain, product or quotient rule.

Examiner Tips and Tricks

  • E(X)=G'X(1) is given in the Formulae Booklet

Worked Example

The probability generating function for a discrete random variable X is given by

GX(t)=181t3(1+2t)4

Find E(X).

ex-of-pgfs

Var(X) of PGFs

How do I find Var(X) of a PGF?

  • The formula is Var(X)=GX''(1)+GX'(1)[GX'(1)]2

  • You may need the chain, product or quotient rule.

  • The first two terms in the formula are E(X2)

    • E(X2)=GX''(1)+GX'(1)

  • The formula comes from 

    • straight G subscript X open parentheses t close parentheses equals straight E open parentheses t to the power of X close parentheses equals sum from blank to blank of space t to the power of x space straight P open parentheses X equals x close parentheses

    • So straight G subscript X apostrophe open parentheses t close parentheses equals stack fraction numerator straight d over denominator straight d t end fraction open square brackets sum from blank to blank of space t to the power of x space straight P open parentheses X equals x close parentheses close square brackets equals sum with blank below and blank on top space x t to the power of x minus 1 end exponent space straight P open parentheses X equals x close parentheses

      • Recall GX'(1)= x P(X=x)=E(X)

    • And straight G subscript X apostrophe apostrophe open parentheses t close parentheses equals stack fraction numerator straight d over denominator straight d t end fraction open square brackets sum from blank to blank of space x t to the power of x minus 1 end exponent space straight P open parentheses X equals x close parentheses close square brackets equals sum with blank below and blank on top space x open parentheses x minus 1 close parentheses t to the power of x minus 2 end exponent space straight P open parentheses X equals x close parentheses

    • Substituting t=1 gives straight G subscript X apostrophe apostrophe open parentheses 1 close parentheses equals sum from blank to blank of space x open parentheses x minus 1 close parentheses space straight P open parentheses X equals x close parentheses space equals space straight E open parentheses X open parentheses X minus 1 close parentheses close parentheses equals space straight E thin space open parentheses X squared close parentheses minus straight E open parentheses X close parentheses

    • Make E(X2) the subject

      • E(X2)=GX''(1)+E(X)

    • Then substitute it into Var(X)=E(X2)(E(X))2

    • And replace E(X) with GX'(1)

Examiner Tips and Tricks

  • Var(X)=GX''(1)+GX'(1)[GX'(1)]2 is given in the Formulae Booklet

Worked Example

The probability generating function for a discrete random variable, X, is given by 

GX(t)=0.2+0.4t+0.3t4+0.1t5

Find Var(X).

varx-of-pgfs

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Mark Curtis

Author: Mark Curtis

Expertise: Maths Content Creator

Mark graduated twice from the University of Oxford: once in 2009 with a First in Mathematics, then again in 2013 with a PhD (DPhil) in Mathematics. He has had nine successful years as a secondary school teacher, specialising in A-Level Further Maths and running extension classes for Oxbridge Maths applicants. Alongside his teaching, he has written five internal textbooks, introduced new spiralling school curriculums and trained other Maths teachers through outreach programmes.

Dan Finlay

Reviewer: Dan Finlay

Expertise: Portfolio Lead

Dan graduated from the University of Oxford with a First class degree in mathematics. As well as teaching maths for over 8 years, Dan has marked a range of exams for Edexcel, tutored students and taught A Level Accounting. Dan has a keen interest in statistics and probability and their real-life applications.