# Variance of two deterministic distributions

#### people

##### New Member
I would like to know the variance of two deterministic distributions when putting them together.

Of course the variance of the two seperate deterministic distributions is zero, right?

I have a machine that processes products and with chance x it processes product A and with chance 1-x it processes product B, so the chances are known.

What will be the variance of the processing time of the machine?

#### people

##### New Member
http://en.wikipedia.org/wiki/Variance#Discrete_random_variable

In particular you may take a look at the Bernoulli distribution which take values [math] \{0, 1\} [/math]:

http://en.wikipedia.org/wiki/Bernoulli_distribution
Thank you I understand it.

My p will be then r/(r+r/t), I can fill that in in the formula for the variance p(1-p). But, in the bernoulli distribution, the values are 0 or 1 and in my assignment this is between different numbers, and I think that these numbers influence the variance. Is that true? So, do I have to convert the variance formula with my other values? And so, how?

#### BGM

##### TS Contributor
Let say you have a discrete random variable [math] X [/math] with only 2 support points [math] \{a, b\} [/math]

And you correctly obtain that [math] \Pr\{X = a\} = p = 1 - \Pr\{X = b\} [/math].

So you look at the formula I posted before. To use it you need to calculate

[math] E[X] = pa + (1 - p)b [/math] first.

Then use either version of it will do.

#### people

##### New Member
Let say you have a discrete random variable [math] X [/math] with only 2 support points [math] \{a, b\} [/math]

And you correctly obtain that [math] \Pr\{X = a\} = p = 1 - \Pr\{X = b\} [/math].

So you look at the formula I posted before. To use it you need to calculate

[math] E[X] = pa + (1 - p)b [/math] first.

Then use either version of it will do.
OK, I have now calculated E [X] andE [X] ^2. And then variance = E [X] ^2 - E [X].
Now I need de covariation coefficient. I wonder if it is simply my E [X]/variance, or something else because it is a discrete distribution?