# variance of random variable formula

$E\left[ {{X}^{2}} \right]=\sum\limits_{x=1}^{m}{{{x}^{2}}f(x)=\frac{1}{m}\sum\limits_{x=1}^{m}{{{x}^{2}}}}$ Problems: a) X i;i= 1;:::;nare independent normal variables with respective parameters i and ˙2 i, then X= P n i=1 X i is normal distribution, show that expectation of Xis n P i=1 i and variance is n i=1 ˙ 2 i. b) A random variable Xwith gamma distribution with parameters (n; );n2N; >0 can be expressed as sum of nindependent exponential random variables: X= Be able to compute the variance and standard deviation of a random variable. Formally, the expected value of a (discrete) random variable X is defined by: The variance of X is defined in terms of the expected value as: From this we can also obtain: Click to share on Twitter (Opens in new window), Click to share on Facebook (Opens in new window). $E\left[ X \right]=\sum\nolimits_{x=1}^{m}{xf(x)=\frac{1}{m}\sum\nolimits_{x=1}^{m}{x}}$ So, for 0 ≤ p ≤ ½ Var(X) is maximum when p = ½           = 1.9, Var(X)    = (0 – 1.9)2x 0.1 + (1 – 1.9)2x 0.2 + (2 – 1.9)2x 0.4 + (3 – 1.9)2x 0.3 An introduction to the concept of the expected value of a discrete random variable. $i.e.,V(X)=\int\limits_{-\infty }^{\infty }{{{\left( x-\mu \right)}^{2}}f(x)dx,\text{Where X is continuous}}$ Variance of a random variable can be defined as the expected value of the square of the difference between the random variable and the mean. $\Rightarrow V(X)=\frac{1}{b-a}\left[ \frac{{{b}^{3}}-{{a}^{3}}}{3} \right]-\frac{{{\left( a+b \right)}^{2}}}{4}$ In the last two sections below, I’m going to give a summary of these derivations. $Variance=V\left( X \right)=E\left[ {{X}^{2}} \right]-E{{\left[ X \right]}^{2}}$ therefore has the nice interpretation of being the probabilty of X taking on the value 1. In the last two sections below, I’m going to give a summary of these derivations. $\Rightarrow E\left( x \right)=(-2)(0.1)+(-1)(0.1)+1(0.2)+2(0.3)+3(0.1)$ $V(X)=E\left[ {{\left( X-\mu \right)}^{2}} \right]$ $\therefore V\left( T \right)={{a}^{2}}V\left( X \right),\left[ as,V(X)=E\left[ {{\left( X-\mu \right)}^{2}} \right] \right]$ The variance of a discrete random variable is given by: $$\sigma^2=\text{Var}(X)=\sum (x_i-\mu)^2f(x_i)$$ The formula means that we take each value of x, subtract the expected value, square that value and multiply that value by … $\therefore V(X)=E\left[ {{X}^{2}} \right]-E{{\left[ X \right]}^{2}}$ $E\left[ {{X}^{2}} \right]=0(p)+1\left( 1-2p \right)+4(p)=1+2p$ The expected value of our binary random variable is. I would love to hear your thoughts and opinions on my articles directly. $f(x)=\frac{1}{m},x=1,2,3,…,m$ $\therefore \text{Standard Deviation}=\sigma =\sqrt{V(X)}=\sqrt{2.64}=1.62$. The variance of that sum or the difference, the variability will increase. $\text{Now mean}=E\left( x \right)=\sum{x.f(x)}$ Taking expectation of both sides, we get $\Rightarrow V(X)=\frac{\left( m+1 \right)\left( 2m+1 \right)}{6}-\frac{{{\left( m+1 \right)}^{2}}}{4}$ For a random variable following this distribution, the expected value is then m 1 = (a + b)/2 and the variance is m 2 − m 1 2 = (b − a) 2 /12. 2. $\therefore V(X)=E\left[ {{X}^{2}} \right]-E{{\left[ X \right]}^{2}}=2p$ Variance of constant is zero, i.e., V (c) = 0 $i.e.,\sigma =\sqrt{V(X)}$ Mean, variance and standard deviation for discrete random variables in Excel. $\therefore V(X)=\frac{{{m}^{2}}-1}{12}$, A random variable X has the following probability function. For the variance of a continuous random variable, the definition is the same and we can still use the alternative formula given by Theorem 3.7.1, only we now integrate to calculate the value: Var ( X) = E [ X 2] − μ 2 = ( ∫ − ∞ ∞ x 2 ⋅ f ( x) d x) − μ 2. Variance of a random variable is discussed in detail here on. 2. Solution: Squaring both sides, we get If you have come this far, it means that you liked what you are reading. In probability and statistics, a random variable, random quantity, aleatory variable, or stochastic variable is described informally as a variable whose values depend on outcomes of a random phenomenon. With that information we can derive the variance of a binary random variate: holds because X can only take on the values zero or one and it holds that and . Cumulant-generating function [ edit ] For n ≥ 2 , the n th cumulant of the uniform distribution on the interval [−1/2, 1/2] is B n / n , where B n is the n th Bernoulli number . I also look at the variance of a discrete random variable. A third use is based on applying the inverse of the probability integral transform to convert random variables from a uniform distribution to have a selected distribution: this is known as inverse transform sampling. Then for what value of p is the Variance of X is maximum. Solution: Probability: (Level 8), Printed from https://nzmaths.co.nz/category/glossary/variance-discrete-random-variable at 7:31am on the 28th November 2020, Learning at home: information for teachers, Subscribe to Variance (of a discrete random variable). The variance of random variable X is often written as Var(X) or σ2 or σ2x. $E\left( X \right)=0(p)+1\left( 1-2p \right)+2(p)=1$ Calculating mean, v Mean, variance and standard deviation for discrete random variables in Excel can be done applying the standard multiplication and sum functions that can be deduced from my Excel screenshot above (the spreadsheet).. Therefore, Why not reach little more and connect with me directly on Facebook, Twitter or Google Plus. The home of mathematics education in New Zealand. therefore, k = 0.1 Let X is a random variable with probability distribution f(x) and mean µ. Understand that standard deviation is a measure of scale or spread. Hence, mean fails to explain the variability of values in probability distribution. Find the value of k and calculate mean, variance and standard deviation. 3. $E\left[ T \right]=E\left[ aX+b \right]$ $i.e.,V(X)=E\left[ {{X}^{2}} \right]+{{\mu }^{2}}-2\mu .\mu =E\left[ {{X}^{2}} \right]-{{\mu }^{2}}$ $P(X=0)=P(X=2)=p$ $\Rightarrow {{\left[ T-E\left[ T \right] \right]}^{2}}={{a}^{2}}{{\left[ X-E\left[ X \right] \right]}^{2}}$ $i.e.,V(X)={{\sum\limits_{x}{\left( x-\mu \right)}}^{2}}f(x),\text{Where X is discrete}$ A measure of spread for a distribution of a random variable that determines the degree to which the values of a random variable differ from the expected value. $\therefore E\left[ {{X}^{2}} \right]=\frac{\left( m+1 \right)\left( 2m+1 \right)}{6}$ Notify me of follow-up comments by email. $E\left( {{X}^{2}} \right)=\int\limits_{a}^{b}{{{x}^{2}}}\frac{1}{b-a}dx=\frac{1}{b-a}\left[ \frac{{{b}^{3}}-{{a}^{3}}}{3} \right]$ The Variance of a random variable is defined as. Save my name, email, and website in this browser for the next time I comment. $f(x)=0,~~otherwise$ Mean of random variables with different probability distributions can have same values. If X is a random variable, then V (aX+b) = a2V (X), where a and b are constants.     = 1.9, E(X2)    = 02 × 0.1 + 12 × 0.2 + 22 × 0.4 + 32 × 0.3 Therefore, variance of random variable is defined to measure the spread and scatter in data. Suppose X is discrete random variable and let the probability mass function of X is given by The square root of the variance is equal to the standard deviation. $\therefore E\left( x \right)=0.8$ The Mean (Expected Value) is: μ = Σxp; The Variance is: Var(X) = Σx 2 p − μ 2… For a discrete random variable the variance is calculated by summing the product of the square of the difference between the value of the random variable and the expected value, and the associated probability of the value of the random variable, taken over all of the values of the random variable. A Random Variable is a variable whose possible values are numerical outcomes of a random experiment. Using Var(X) = E(X2) – [E(X)]2, E(X)    = 0 x 0.1 + 1 x 0.2 + 2 x 0.4 + 3 x 0.3 $\therefore V\left( aX+b \right)={{a}^{2}}V\left( X \right)$ $\Rightarrow T-E\left[ T \right]=aX+b-E\left[ aX+b \right]$ So, let’s calculate $V(X)=\frac{{{\left( b-a \right)}^{2}}}{12}$, Your email address will not be published. We know that, Σ f(x) = 1 It is denoted as σ. Find the variance of X. Hence, mean fails to explain the variability of values in probability distribution. Variance of a random variable is discussed in detail here on. Your email address will not be published. MAKAUT BCA 1ST Semester Previous Year Question Papers 2018 | 2009 | 2010 | 2011 | 2012, Abstract Algebra – Group, Subgroup, Abelian group, Cyclic group, Iteration Method or Fixed Point Iteration – Algorithm, Implementation in C With Solved Examples, Theory of Equation – Descartes’ Rule of Signs With Examples. Let the random variable X have the distribution:

$E\left[ {{X}^{2}} \right]=\sum\limits_{x=1}^{m}{{{x}^{2}}f(x)=\frac{1}{m}\sum\limits_{x=1}^{m}{{{x}^{2}}}}$ Problems: a) X i;i= 1;:::;nare independent normal variables with respective parameters i and ˙2 i, then X= P n i=1 X i is normal distribution, show that expectation of Xis n P i=1 i and variance is n i=1 ˙ 2 i. b) A random variable Xwith gamma distribution with parameters (n; );n2N; >0 can be expressed as sum of nindependent exponential random variables: X= Be able to compute the variance and standard deviation of a random variable. Formally, the expected value of a (discrete) random variable X is defined by: The variance of X is defined in terms of the expected value as: From this we can also obtain: Click to share on Twitter (Opens in new window), Click to share on Facebook (Opens in new window). $E\left[ X \right]=\sum\nolimits_{x=1}^{m}{xf(x)=\frac{1}{m}\sum\nolimits_{x=1}^{m}{x}}$ So, for 0 ≤ p ≤ ½ Var(X) is maximum when p = ½           = 1.9, Var(X)    = (0 – 1.9)2x 0.1 + (1 – 1.9)2x 0.2 + (2 – 1.9)2x 0.4 + (3 – 1.9)2x 0.3 An introduction to the concept of the expected value of a discrete random variable. $i.e.,V(X)=\int\limits_{-\infty }^{\infty }{{{\left( x-\mu \right)}^{2}}f(x)dx,\text{Where X is continuous}}$ Variance of a random variable can be defined as the expected value of the square of the difference between the random variable and the mean. $\Rightarrow V(X)=\frac{1}{b-a}\left[ \frac{{{b}^{3}}-{{a}^{3}}}{3} \right]-\frac{{{\left( a+b \right)}^{2}}}{4}$ In the last two sections below, I’m going to give a summary of these derivations. $Variance=V\left( X \right)=E\left[ {{X}^{2}} \right]-E{{\left[ X \right]}^{2}}$ therefore has the nice interpretation of being the probabilty of X taking on the value 1. In the last two sections below, I’m going to give a summary of these derivations. $\Rightarrow E\left( x \right)=(-2)(0.1)+(-1)(0.1)+1(0.2)+2(0.3)+3(0.1)$ $V(X)=E\left[ {{\left( X-\mu \right)}^{2}} \right]$ $\therefore V\left( T \right)={{a}^{2}}V\left( X \right),\left[ as,V(X)=E\left[ {{\left( X-\mu \right)}^{2}} \right] \right]$ The variance of a discrete random variable is given by: $$\sigma^2=\text{Var}(X)=\sum (x_i-\mu)^2f(x_i)$$ The formula means that we take each value of x, subtract the expected value, square that value and multiply that value by … $\therefore V(X)=E\left[ {{X}^{2}} \right]-E{{\left[ X \right]}^{2}}$ $E\left[ {{X}^{2}} \right]=0(p)+1\left( 1-2p \right)+4(p)=1+2p$ The expected value of our binary random variable is. I would love to hear your thoughts and opinions on my articles directly. $f(x)=\frac{1}{m},x=1,2,3,…,m$ $\therefore \text{Standard Deviation}=\sigma =\sqrt{V(X)}=\sqrt{2.64}=1.62$. The variance of that sum or the difference, the variability will increase. $\text{Now mean}=E\left( x \right)=\sum{x.f(x)}$ Taking expectation of both sides, we get $\Rightarrow V(X)=\frac{\left( m+1 \right)\left( 2m+1 \right)}{6}-\frac{{{\left( m+1 \right)}^{2}}}{4}$ For a random variable following this distribution, the expected value is then m 1 = (a + b)/2 and the variance is m 2 − m 1 2 = (b − a) 2 /12. 2. $\therefore V(X)=E\left[ {{X}^{2}} \right]-E{{\left[ X \right]}^{2}}=2p$ Variance of constant is zero, i.e., V (c) = 0 $i.e.,\sigma =\sqrt{V(X)}$ Mean, variance and standard deviation for discrete random variables in Excel. $\therefore V(X)=\frac{{{m}^{2}}-1}{12}$, A random variable X has the following probability function. For the variance of a continuous random variable, the definition is the same and we can still use the alternative formula given by Theorem 3.7.1, only we now integrate to calculate the value: Var ( X) = E [ X 2] − μ 2 = ( ∫ − ∞ ∞ x 2 ⋅ f ( x) d x) − μ 2. Variance of a random variable is discussed in detail here on. 2. Solution: Squaring both sides, we get If you have come this far, it means that you liked what you are reading. In probability and statistics, a random variable, random quantity, aleatory variable, or stochastic variable is described informally as a variable whose values depend on outcomes of a random phenomenon. With that information we can derive the variance of a binary random variate: holds because X can only take on the values zero or one and it holds that and . Cumulant-generating function [ edit ] For n ≥ 2 , the n th cumulant of the uniform distribution on the interval [−1/2, 1/2] is B n / n , where B n is the n th Bernoulli number . I also look at the variance of a discrete random variable. A third use is based on applying the inverse of the probability integral transform to convert random variables from a uniform distribution to have a selected distribution: this is known as inverse transform sampling. Then for what value of p is the Variance of X is maximum. Solution: Probability: (Level 8), Printed from https://nzmaths.co.nz/category/glossary/variance-discrete-random-variable at 7:31am on the 28th November 2020, Learning at home: information for teachers, Subscribe to Variance (of a discrete random variable). The variance of random variable X is often written as Var(X) or σ2 or σ2x. $E\left( X \right)=0(p)+1\left( 1-2p \right)+2(p)=1$ Calculating mean, v Mean, variance and standard deviation for discrete random variables in Excel can be done applying the standard multiplication and sum functions that can be deduced from my Excel screenshot above (the spreadsheet).. Therefore, Why not reach little more and connect with me directly on Facebook, Twitter or Google Plus. The home of mathematics education in New Zealand. therefore, k = 0.1 Let X is a random variable with probability distribution f(x) and mean µ. Understand that standard deviation is a measure of scale or spread. Hence, mean fails to explain the variability of values in probability distribution. Find the value of k and calculate mean, variance and standard deviation. 3. $E\left[ T \right]=E\left[ aX+b \right]$ $i.e.,V(X)=E\left[ {{X}^{2}} \right]+{{\mu }^{2}}-2\mu .\mu =E\left[ {{X}^{2}} \right]-{{\mu }^{2}}$ $P(X=0)=P(X=2)=p$ $\Rightarrow {{\left[ T-E\left[ T \right] \right]}^{2}}={{a}^{2}}{{\left[ X-E\left[ X \right] \right]}^{2}}$ $i.e.,V(X)={{\sum\limits_{x}{\left( x-\mu \right)}}^{2}}f(x),\text{Where X is discrete}$ A measure of spread for a distribution of a random variable that determines the degree to which the values of a random variable differ from the expected value. $\therefore E\left[ {{X}^{2}} \right]=\frac{\left( m+1 \right)\left( 2m+1 \right)}{6}$ Notify me of follow-up comments by email. $E\left( {{X}^{2}} \right)=\int\limits_{a}^{b}{{{x}^{2}}}\frac{1}{b-a}dx=\frac{1}{b-a}\left[ \frac{{{b}^{3}}-{{a}^{3}}}{3} \right]$ The Variance of a random variable is defined as. Save my name, email, and website in this browser for the next time I comment. $f(x)=0,~~otherwise$ Mean of random variables with different probability distributions can have same values. If X is a random variable, then V (aX+b) = a2V (X), where a and b are constants.     = 1.9, E(X2)    = 02 × 0.1 + 12 × 0.2 + 22 × 0.4 + 32 × 0.3 Therefore, variance of random variable is defined to measure the spread and scatter in data. Suppose X is discrete random variable and let the probability mass function of X is given by The square root of the variance is equal to the standard deviation. $\therefore E\left( x \right)=0.8$ The Mean (Expected Value) is: μ = Σxp; The Variance is: Var(X) = Σx 2 p − μ 2… For a discrete random variable the variance is calculated by summing the product of the square of the difference between the value of the random variable and the expected value, and the associated probability of the value of the random variable, taken over all of the values of the random variable. A Random Variable is a variable whose possible values are numerical outcomes of a random experiment. Using Var(X) = E(X2) – [E(X)]2, E(X)    = 0 x 0.1 + 1 x 0.2 + 2 x 0.4 + 3 x 0.3 $\therefore V\left( aX+b \right)={{a}^{2}}V\left( X \right)$ $\Rightarrow T-E\left[ T \right]=aX+b-E\left[ aX+b \right]$ So, let’s calculate $V(X)=\frac{{{\left( b-a \right)}^{2}}}{12}$, Your email address will not be published. We know that, Σ f(x) = 1 It is denoted as σ. Find the variance of X. Hence, mean fails to explain the variability of values in probability distribution. Variance of a random variable is discussed in detail here on. Your email address will not be published. MAKAUT BCA 1ST Semester Previous Year Question Papers 2018 | 2009 | 2010 | 2011 | 2012, Abstract Algebra – Group, Subgroup, Abelian group, Cyclic group, Iteration Method or Fixed Point Iteration – Algorithm, Implementation in C With Solved Examples, Theory of Equation – Descartes’ Rule of Signs With Examples. 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