# expected value and variance

(x1 - E [X])^2, p (x2). P(X = 2) = 1/6 (the probability that you throw a 2 is 1/6) The variance of a discrete random variable is given by: σ 2 = Var (X) = ∑ (x i − μ) 2 f (x i) The formula means that we take each value of x, subtract the expected value, square that value and multiply that value by its probability. $$\textrm{Var}(X)=E\big[(X-\mu_X)^2\big]=EX^2-(EX)^2.$$ So, how do we use the concept of expected value to calculate the mean and variance of a probability distribution? Now that we can find what value we should expect, (i.e. $$EX=\sum_{x_k \in R_X} x_k P_X(x_k).$$ Variance measures the difference from the expected value. the expected value), it is also of interest to give a measure of the variability. In my post on expected value, I defined it to be the sum of the products of each possible value of a random variable and that value’s probability.. For a discrete random variable X, the variance of X is written as Var(X). If Xis a random variable with values x 1;x 2;:::;x n, corresponding probabilities p 1;p 2;:::;p n, and expected value = E(X), then Variance = ˙ 2(X) = p 1(x 1 2 ) 2 +p and 2(x 2 ) + +p n(x n ) Standard Deviation = ˙(X) = p Variance : P(X = 5) = 1/6 (the probability that you throw a 5 is 1/6) If you think about it, 3.5 is halfway between the possible values the die can take and so this is what you should have expected. Var(X) = E[ (X – m)2 ]            where m is the expected value E(X). P(X = 6) = 1/6 (the probability that you throw 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 to find its expected value, we can write, $E[aX+b]=aEX+b$, for all $a,b \in \mathbb{R}$, and. \end{equation} . P(X = 4) = 1/6 (the probability that you throw a 4 is 1/6) 2x & \quad 0 \leq x \leq 1\\ $E[X_1+X_2+...+X_n]=EX_1+EX_2+...+EX_n$, for any set of random variables $X_1, X_2,...,X_n$. $$\hspace{70pt} E[g(X)]=\sum_{x_k \in R_X} g(x_k)P_X(x_k) \hspace{70pt} (4.2)$$ 3.2.1 - Expected Value and Variance of a Discrete Random Variable . Expected Value (or EV) is a measure of what you can expect to win or lose per bet placed in the long run. Expected Value, Mean, and Variance Using Excel This tutorial will calculate the mean and variance using an expected value. = a2E(X2) - a2E2(X) = a2Var(X). In particular, usually summations are replaced by integrals and PMFs are replaced by PDFs. Var [X] = sum (p (x1). Then we'll discuss properties of expected value and variance with respect to arithmetic operations and introduce measures of independence between random variables. The expected value of X is usually written as E(X) or m. So the expected value is the sum of: [(each of the possible outcomes) × (the probability of the outcome occurring)]. Expectation and Variance The expected value (or mean) of X, where X is a discrete random variable, is a weighted average of the possible values that X can take, each value being weighted according to the probability of that event occurring. 0 & \quad x < a \textrm{ or } x > b The expected value (or mean) of X, where X is a discrete random variable, is a weighted average of the possible values that X can take, each value being weighted according to the probability of that event occurring. It turns out (and we Linearity of expected value 6:56. 0 & \quad \text{otherwise} The expected value of a constant is just the constant, so for example E(1) = 1. (xn - E [X])^2) 1 Linear transformations of random variables 8:51. Expected Value is without variance. formula for the variance of a random variable. The proofs and ideas are very analogous to the discrete case, so sometimes we state the results without There are six possible outcomes: 1, 2, 3, 4, 5, 6. (x2 - E [X])^2,..., p (x1). In fact: You is because: Var[aX + b] = E[ (aX + b)2 ] - (E [aX + b])2 . random variables. \begin{array}{l l} Remember that the expected value of a discrete random variable can be obtained as Expected value of product of independent random variables with same expected value and variance 0 Find variance and general formula for for r$^{th}$ moment for random variable uniform over (0,1) = E[ a2X2 + 2abX + b2] - (aE(X) + b)2 As we mentioned earlier, the theory of continuous random variables is very similar to the theory of discrete Let X represent the outcome of the experiment. \nonumber f_X(x) = \left\{ = a2E(X2) + 2abE(X) + b2 - a2E2(X) - 2abE(X) - b2 In this example, Harrington Health Food stocks 5 loaves of Neutro-Bread. The standard deviation of X is the square root of Var(X). Properties of Expected values and Variance Christopher Croke University of Pennsylvania Math 115 UPenn, Fall 2011 Christopher Croke Calculus 115. P(X = 3) = 1/6 (the probability that you throw a 3 is 1/6) Variance of a Discrete Random Variable . (How far a set of numbers are spread out from their average value.) mathematical derivations for the purpose of brevity. Probability distributions, including the t-distribution, have several moments, including the expected value, variance, and standard deviation (a moment is a summary measure of a probability distribution): The first moment of a distribution is the expected value, E (X), which represents the mean or average value of the distribution. Moreover, it determines the degree to which the values of a random variable differ from the expected value. Multiplying a random variable by a constant multiplies the expected value by that constant, so E[2X] = 2E[X]. As a result, it’s defined with We have (a) $VarX=E[X^{2}]-(EX)^{2}$ (b)$Var(aX+b)=a^{2}VarX$ For calculating variance in given problems we will mostly use (a). It turns out (and we Y = X2 + 3 so in this case r(x) = x2 + 3. $$\hspace{70pt} E[g(X)]=\int_{-\infty}^{\infty} g(x) f_X(x) dx \hspace{70pt} (4.3)$$, $=\frac{1}{b-a} \bigg[ \frac{1}{2}x^2 \bigg]_{a}^{b} dx$, $= \bigg[\frac{1}{n+2}x^{n+2}+\frac{1}{2(n+1)}x^{n+1} \bigg]_{0}^{1}$, $$=E\big[(X-\mu_X)^2\big]=\int_{-\infty}^{\infty} (x-\mu_X)^2 f_X(x)dx$$, $$=EX^2-(EX)^2=\int_{-\infty}^{\infty} x^2 f_X(x)dx-\mu_X^2$$, $= \bigg[-\frac{3}{2}x^{-2} \bigg]_{1}^{\infty}$, As we saw, the PDF of $X$ is given by \frac{1}{b-a} & \quad a < x < b\\ Since most of the statistical quantities we are studying will be averages it is very important you know where these formulas come from. \begin{array}{l l} \begin{equation} Properties of Expected values and Variance Christopher Croke University of Pennsylvania Math 115 UPenn, Fall 2011 Christopher Croke Calculus 115. a continuous random variable as. Expected value Consider a random variable Y = r(X) for some function r, e.g. Real value is your actual result (profit), which depends on the outcome of the matches. continuous random variables. \end{equation} Then sum all of those values. Therefore P(X = 1) = 1/6 (this means that the probability that the outcome of the experiment is 1 is 1/6) Remember the law of the unconscious statistician (LOTUS) for discrete random variables: So the expectation is 3.5 . 4.1.2 Expected Value and Variance As we mentioned earlier, the theory of continuous random variables is very similar to the theory of discrete random variables. Expected value Consider a random variable Y = r(X) for some function r, e.g. \end{array} \right. The variance of a random variable tells us something about the spread of the possible values of the variable. EXPECTED VALUE AND VARIANCE n = 100 n = 10000 Winning Frequency Relative Frequency Relative Frequency Frequency 1 17.17 1681.1681 -2 17.17 1678.1678 3 16.16 1626.1626 -4 18.18 1696.1696 5 16.16 1686.1686 -6 16.16 1633.1633 Table 6.1: Frequencies for dice game. Let $X$ be a continuous random variable with PDF EN FR
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