Use MathJax to format equations. The answer is given and should be 3/8. The correct way to proceed with (1) is to multiply out $Y(2Y-X)$ to obtain $2Y^2-XY$. Why did mainframes have big conspicuous power-off buttons? This wasn't a coincidence -- it would have happened if the 200 was 1000, 10 million, or 13,798,235,114. Give the number of the probability of success and values of x, expected value calculator will notify you about the expected value for a discrete random variable. You can also use our other calculators. That is to say, To see this, note Additionally, I have been given a discrete random variable Y, which is independent of X, and has probability function py(y) = 3/4 if y = 0, 1/4 if y = 1, 0 otherwise. A discrete random variable is finite if its list of possible values has a fixed (finite) number of elements in it (for example, the number of smoking ban supporters in a random sample of 100 voters has to be between 0 and 100). If covariance is positive, then increasing one variable results in the increase of another variable. Enter probability or weight and data number in each row: rev 2020.11.24.38066, The best answers are voted up and rise to the top, Mathematics Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us. E[Y] = \frac34(0) + \frac14(1) = \frac14\\ Provide this information, the calculation is very simple. $$\begin{array}{rcl} Now with your probability distribution as given: The expected value associated with a discrete random variable $X$, denoted by either $E(X)$ or $\mu$ (depending on context) is the theoretical mean of $X$. This is because any events happenings probabilities can’t be greater than 100%. Thus, I know the definition of covariance and I'm trying to solve some exercises. Is it too late for me to get into competitive chess? Functions: What They Are and How to Deal with Them, Normal Probability Calculator for Sampling Distributions, Calculator of Mean And Standard Deviation for a Probability Distribution, Mean and Standard Deviation of Discrete Distribution. But think about where these numbers came from -- we could write instead: We can then factor out a 200 from every term in the numerator, which would cancel with the 200 in the denominator, yielding. E(Y^2) = \frac34(0^2) + \frac14(1^2) = \frac14\\ For a discrete probability, the … Also, you can understand how the algorithm is used by a calculator to find the discrete random variable’s expected value. It is easy to learn to find the expected value. E[YX]=\frac34\frac12(0\cdot (-1)) + \frac34\frac14(0\cdot (0)) + \frac34\frac14(0\cdot (1)) \\+ \frac14\frac12(1\cdot (-1)) + \frac14\frac14(1\cdot (0)) + \frac14\frac14(1\cdot (1)) = -\frac18+\frac1{16} = Since X and Y are independent, I should be able to use the rule E(X * Y) = E(X) * E(Y) so that I get: Cov(Y, 2Y-X) = E[Y] * E[2Y - X] - E[Y] * E[2Y - X]. For calculating single discrete random variables of Expected Value, one must multiply the value of the variable by the probability of that value occurring. How does this Poisson distribution calculator work? To subscribe to this RSS feed, copy and paste this URL into your RSS reader. &=& c \displaystyle{\sum_{x \in S_x} \left[ x \cdot P(x) \right]}\\\\ Population and sampled standard deviation calculator. However, it's not true that $Y$ and $2Y-X$ are independent, so you cannot say With this in mind, and assuming that this random variable has an outcome/sample space of $S$ and probability mass function $P$, this expected value is given by. Discrete random variables can take on either a finite or at most a countably infinite set of discrete values (for example, the integers). &=& c \cdot E(X) Most importantly this value is the variables long-term average value. For example when I try to deduce from the definition Cov(X, Y) = E[(X - E[X])(Y - E[Y])] I get the following: Cov(Y, 2Y-X) = E[(Y - E[Y])((2Y - X) - E[2Y - X])]. \displaystyle{\sum_{x \in S_x,\,y \in S_y} x \cdot P(X=x \textrm{ and } Y=y)} &=& \displaystyle{\sum_{x \in S_x} \left[ \sum_{y \in S_y} x \cdot P(X=x \textrm{ and } Y=y) \right]}\\\\