The expected value (or mean) the X, whereby X is a discrete random variable, is a weighted average of the feasible values the X have the right to take, each worth being weighted follow to the probability the that event occurring. The meant value of X is generally written as E(X) or m.
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E(X) = S x P(X = x)
So the intended value is the sum of: <(each the the possible outcomes) × (the probability that the result occurring)>.
In more concrete terms, the expectation is what girlfriend would suppose the outcome of one experiment to it is in on average.
What is the intended value as soon as we role a fair die?
There space six feasible outcomes: 1, 2, 3, 4, 5, 6. Every of these has a probability of 1/6 that occurring. Permit X stand for the outcome of the experiment.
Therefore P(X = 1) = 1/6 (this means that the probability the the result of the experiment is 1 is 1/6)P(X = 2) = 1/6 (the probability the you throw a 2 is 1/6)P(X = 3) = 1/6 (the probability the you throw a 3 is 1/6)P(X = 4) = 1/6 (the probability that you throw a 4 is 1/6)P(X = 5) = 1/6 (the probability that you throw a 5 is 1/6)P(X = 6) = 1/6 (the probability that you litter a 6 is 1/6)
E(X) = 1×P(X = 1) + 2×P(X = 2) + 3×P(X = 3) + 4×P(X=4) + 5×P(X=5) + 6×P(X=6)
Therefore E(X) = 1/6 + 2/6 + 3/6 + 4/6 + 5/6 + 6/6 = 7/2
So the expectation is 3.5 . If you think about it, 3.5 is halfway between the possible values the die deserve to take and also so this is what girlfriend should have expected.
Expected worth of a role of X
To find E< f(X) >, wherein f(X) is a function of X, usage the following formula:
E< f(X) > = S f(x)P(X = x)
For the above experiment (with the die), calculate E(X2)
Using our notation above, f(x) = x2
f(1) = 1, f(2) = 4, f(3) = 9, f(4) = 16, f(5) = 25, f(6) = 36P(X = 1) = 1/6, P(X = 2) = 1/6, etc
So E(X2) = 1/6 + 4/6 + 9/6 + 16/6 + 25/6 + 36/6 = 91/6 = 15.167
The intended value that a continuous is simply the constant, therefore for example E(1) = 1. Multiplying a random variable through a continuous multiplies the expected value by the constant, therefore E<2X> = 2E
A advantageous formula, where a and b room constants, is:
The variance of a arbitrarily variable tells us something around the spread of the feasible values of the variable. Because that a discrete arbitrarily variable X, the variance of X is composed as Var(X).
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Var(X) = E< (X – m)2 > where m is the intended value E(X)
This can additionally be created as:
Var(X) = E(X2) – m2
The conventional deviation of X is the square source of Var(X).
Note the the variance does no behave in the same way as expectation once we main point and include constants to arbitrarily variables. In fact:
You is because: Var
= E< a2X2 + 2abX + b2> - (aE(X) + b)2= a2E(X2) + 2abE(X) + b2 - a2E2(X) - 2abE(X) - b2= a2E(X2) - a2E2(X) = a2Var(X)