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Question: Write the formula for \[E(X)\] and \(Var(X)\)...

Write the formula for E(X)E(X) and Var(X)Var(X)

Explanation

Solution

First, we know the random variance of the random variable X is the mean or expected value of the square deviation from the mean of X.
Using their definition, we can arrive at a simpler formula for variance, both for continuous and discrete variables.
These formulas are mostly used in the statistics for the distribution method.

Complete step-by-step answer:
We know that the variance measures how far a set of numbers is spread out, from their average value.
Also, we are aware that variance is the square of standard deviation.
The variance of random variable X is represented by Var(X)Var(X)
Using this definition for the variance for the random variable X, we can write the variance as Var(X)=E[(X−E[X])2]Var(X) = E[{(X - E[X])^2}], where E(X) represents the expected value or mean for the random variable X.
We can expand the equation as, Var(X)=E[(X−E[X])2]⇒E[X2−2X.E[X]+E[X]2]Var(X) = E[{(X - E[X])^2}] \Rightarrow E[{X^2} - 2X.E[X] + E{[X]^2}]
By the use of (a−b)2{(a - b)^2}formula,
Now since giving the expectation values inside the equation we get, Var(X)=E[X2]−2E[X].E[X]+E[X]2Var(X) = E[{X^2}] - 2E[X].E[X] + E{[X]^2}
Further solving this we get, Var(X)=E[X2]−E[X]2Var(X) = E[{X^2}] - E{[X]^2} (since E[X]E[X]or mm)
Because the expected value of X is usually written as E[X]E[X]ormm.
Thus, we get Var(X)=E[X2]−E[X]2⇒Var(X)=E[X2]−m2Var(X) = E[{X^2}] - E{[X]^2} \Rightarrow Var(X) = E[{X^2}] - {m^2}
For finding the expected value of X, the discrete random variable is known as X, is a weighted average of the possible values that X can take that each value from weighted probability from according to that event occurring.

The formula is E[X]=Sf(x)×P(X=x)E[X] = Sf(x) \times P(X = x)which is the expected value E(X) in a discrete random variable.
The expected value can be also expressed as E(X)=∑i=1nXiP(Xi)E(X) = \sum\limits_{i = 1}^n {{X_i}} P({X_i})
Thus, the formula can be rewritten as ∑i=1nXiP(Xi)=Sf(x)×P(X=x)\sum\limits_{i = 1}^n {{X_i}} P({X_i}) = Sf(x) \times P(X = x)
and variance of X is Var(X)=E[X2]−m2Var(X) = E[{X^2}] - {m^2}

Note: So, the expected value is the sum of each possible outcome into times of the probability of the outcome occurring in the expected value.
We also say that the variance of X equal to the difference of the mean of the square of X and the square of the mean of X for the variance of the X.
The variance value is interrelated to the given expected value for the distribution.
For the continuous random variable for expected value is μ=E(X)⇒∫−∞∞xfX(x)dx\mu = E(X) \Rightarrow \int\limits_{ - \infty }^\infty {x{f_X}} (x)dxand variance is Var(X)=E(X−μ)2Var(X) = E{(X - \mu )^2}