sums of independent random variables
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Statlect The Digital Textbook
Index>Additional topics in probability theory
Sums of independent random variables
This lecture discusses how to derive the distribution of the sum of two independent random variables.We explain first
how to derive the distribution functionof the sum and then how to derive its probability mass function(if the summands
are discrete) or its probability density function(if the summands are continuous).
Distribution function of a sum
The following proposition characterizes the distribution function of the sum in terms of the distribution functions of the
two summands:
Proposition Let and be two independent random variables and denote by and their
distribution functions. Let:
and denote the distribution function of by . The following holds:
or:
Proof
Example Let be a uniform random variablewith support and probability density function:
and another uniform random variable, independent of , with support and probability density
function:
The distribution function of is:
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The distribution function of is:
There are four cases to consider:
1. If , then:
2. If , then:
3. If , then:
4. If , then:
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Combining these four possible cases, we obtain:
Probability mass function of a sum
When the two summands are discrete random variables, the probability mass function of their sum can be derived as
follows:
Proposition Let and be two independent discrete random variables and denote by and their
respective probability mass functions and by and their supports. Let:
and denote the probability mass function of by . The following holds:
or:
Proof
The two summations above are called convolutions (of two probability mass functions).
Example Let be a discrete random variable with support and probability mass function:
and another discrete random variable, independent of , with support and probability mass
function:
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Define
Its support is:
The probability mass function of , evaluated at is:
Evaluated at , it is:
Evaluated at , it is:
Therefore, the probability mass function of is:
Probability density function of a sum
When the two summands are absolutely continuous random variables, the probability density function of their sum can
be derived as follows:
Proposition Let and be two independent absolutely continuous random variables and denote by and
their respective probability density functions. Let:
and denote the probability density function of by . The following holds:
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or:
Proof
The two integrals above are called convolutions (of two probability density functions).
Example Let be an exponential random variablewith support and probability density function:
and another exponential random variable, independent of , with support and probability density
function:
Define:
The support of is:
When , the probability density function of is:
Therefore, the probability density function of is:
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Featured pagesStudent t distribution
Gamma function
Normal distribution
Main sectionsMathematical tools
Fundamentals of probability
Additional topics in probability
Glossary entriesAlternative hypothesis
Convolutions
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More details
Sum of n independent random variables
We have discussed above how to derive the distribution of the sum of two independent random variables. How do we
derive the distribution of the sum of more than two mutually independent random variables? Suppose , , ...,
are mutually independent random variables and let be their sum:
The distribution of can be derived recursively, using the results for sums of two random variables given above:
1. first, define:
and compute the distribution of ;
2. then, define:
and compute the distribution of ;
3. and so on, until the distribution of can be computed from:
Solved exercises
Below you can find some exercises with explained solutions:
1. Exercise set 1
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