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### Mean and Variance of Distributions

```
Date: 12/07/2000 at 09:37:18
From: Vaughn Duff
Subject: Mean and Variance of Distributions

I'm stuck on the following problems. Any insight into deriving a
solution would be greatly appreciated. Thank you!

I am trying to derive the expected value (mean) and variance for these
distributions:

1) Exponential with parameter 1/k:

f(w) = (k)*e^(-w*k) or maybe it should be f(w)=(1/k)*e^(-w/k)

The expected value is equal to:

SUM[f(w)*w]     (discrete case)
or
inf
INT[f(w)*w dw]   (continuous case)
0

The variance is equal to:

SUM[f(w)*w^2 dw - (mean)^2]

The derivation should be equal to:

E(w) = 1/k var(w)
= 1/k^2

2) Hypergeometic distribution: define y = the number of S's in a
random sample of size n (taken without replacement) from a population
of N_s S's and N_f F's:

p(y) = ((N_s C y)*(N_f C n-y))/(N C n)

The definition of the mean and variance are the same as in the first
case. The derivation should be equal to:

E(y) = n*(N_s/N) var(y)
= n*(N_s/N)*(1-N_s/N)*((N-n)/(N-1))
```

```
Date: 12/07/2000 at 11:03:04
From: Doctor Anthony
Subject: Re: Mean and Variance of Distributions

For the first case, show that for a Poisson process the time interval
between events has an exponential distribution.

If the average is k per unit time, then in t units of time the average
would be kt.

The probability of no event P(0) in time t is therefore e^(-kt).

The probability P(T>t) = prob. no events in time 0 -> t = e^(-kt), so

P(T<t) = 1 - e^(-kt)

This is the c.d.f. of the time to first event.

To get the p.d.f. we differentiate this and have:

f(t) = ke^(-kt)

This is pdf of time interval between events, and is the exponential
distribution.

inf
E(t) = INT[t.ke^(-kt).dt]
0

and integrating by parts

= t(-1/k)ke^(-kt) + INT[e^(-kt).dt]

= 0  - (1/k)[e^(-kt)] from 0 to infinity

= -(1/k)[0 - 1]

E(t) = 1/k

To get variance we first find:

E(t^2) = INT[t^2.(ke^(-kt).dt]

and integrating by parts (twice) gives 2/k^2 and so

var(t) = 2/k^2 - 1/k^2

=  1/k^2

For the second case, suppose we have a population of b black and g
green elements. A sample of size r is taken (without replacement) and
we wish to find the expected number of black elements in the sample.

We could calculate this by direct methods but a better way is to
introduce a new variable x(k) which takes the value 1 or 0 depending
on whether the kth element in the sample is or is not black.

By symmetry this probability for each value of k is b/(b+g)

The expected number of black elements for each k is:

E[x(k)] = 1.b/(b+g) + 0.g/(b+g)
= b/(b+g)

E[b] = E[x(1)] + E[x(2)] + ... + E[x(r)]

and so

E[b] = b/(b+g) + b/(b+g) + ... + b/(b+g)

to r terms giving:

E[b] = rb/(b+g)

Variance
--------
To calculate the variance we continue as follows:

E(x(k)^2) = 1^2.b/(b+g) + 0.g/(b+g)
= b/(b+g)

Var(x(k)) = b/(b+g) - b^2/(b+g)^2

b(b+g) - b^2
= ------------
(b+g)^2

bg
=  -------
(b+g)^2

Now we must add together the means and variances of the individual
results of the x(i)'s. To do this we first require the covariance of
x(j).x(k)

Covariance
-----------
x(j).x(k) = 1 if BOTH x(j) and x(k) = 1, otherwise x(j).x(k) = 0.

Probability x(j).x(k) = 1 is given by:

b     b-1
--- . -----
b+g   b+g-1

and so

E[x(j).x(k)] = b(b-1)/[(b+g)(b+g-1)]

COV[x(j).x(k)] = E[x(j).x(k)] - E(x(j)).E(x(k))

= b(b-1)/[(b+g)(b+g-1)] - [b/(b+g)].[b/(b+g)]

b(b-1)(b+g) - b^2(b+g-1)
= ------------------------
(b+g)^2.(b+g-1)

b^2(b+g) - b(b+g) - b^2(b+g) + b^2
= ----------------------------------
(b+g)^2.(b+g-1)

-b^2 - bg + b^2
=  ---------------
(b+g)^2.(b+g-1)

-bg
=  ---------------
(b+g)^2.(b+g-1)

And so finally for a sample size r taken from the population of b+g we
get:

rb
E[S(r)] = ---
b+g

The covariance of x(j) with all other x(i) gives r-1 of these terms,
so:

rbg          r-1
Var[S(r)] = -------  [1 - -----]
(b+g)^2       b+g-1

- Doctor Anthony, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
College Statistics

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