ransmission errors occur randomly and independently in time and can be modelled as a
Poisson random variable. Let X = the number of transmission errors per unit time, assume:
X ~ Pois (λ = 10).
However not all transmission errors are detected. Suppose that a transmission error gets detected
with probability 0.6 (independent of all other transmissions). Let Y = the number of
transmission errors detected. Given X transmission errors were sent, Y|X ~ binomial (n = X, p =
.6).
Ultimately, we are interested in the number of transmission errors sent, however this is
impossible to observe. Rather, we can only "observe" transmission errors that are detected; thus
we are interested in the distribution of X|Y.
In the following, we will estimate the probability distribution of X given that Y = 8 transmission
errors were detected.
For a large number of trials (N = 1000000), lets simulate X, the number of transmission errors
sent, and Y, (for each X) the number of transmission errors detected.
>> N = 1000000;
>> X = random(‘pois’,10,[1,N]): % creates of vector of length N containing randomly
generated values from a Pois (λ = 10) distribution.
>> i = 1:N;
>> Y(i) = random(‘bino’, X(i), 0.6); % creates of vector of length N containing randomly
generated values from a binomial distribution with n = X(i) trials and p=
P(success).
Now, let’s condition of Y = 8 (8 transmission errors were detected) and extract from vector X, all
of the times when the # or transmission errors sent resulted in 8 transmission errors detected. I
will name this new vector XgY8 (short for X given Y = 8). Here is code to extract entries that
meet a certain condition from a vector:
>> position = find(Y==8); % use the double = , to test equality at the component level.
>> XgY8 = X(position); % creates a new vector containing X-values only if they
resulted in Y = 8.
WHAT TO SOLVE FOR
1. Provide a well labelled, normalized, histogram of XgY8 (the distribution of transmission
errors sent GIVEN 8 transmission errors were detected)
Note: XgY8 is a discrete random variable, so to normalize, use
"normalization","probability"
in your histogram command.
2. Calculate the mean of XgY8. This provides an estimate for E(X|Y=8); i.e. the expected
number of transmission errors sent given 8 errors are detected.
