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Feb 9, 2020 · Let X and Y be discrete random variables. Prove the linearity of expectation E(X+Y) = E(X) + E(Y). An exercise problem in probability theory. The solution is given.
Jan 29, 2019 · Let T T denote the set of products of two elements in S S. Then XY X Y is a discrete random variable taking values in the set T T. Now suppose I ask you what the probability of XY X Y taking the value z z is. Well clearly if X = x X = x, then Y Y must equal z/x z / x.
Solution. Verified by Toppr. We have , yx =ey−x. Taking log both side. ⇒ logyx =logey−x. ⇒ xlogy =y−x.......(1) Differentiating w.r.t x, we get. 1.logy+x 1 ydy dx = dy dx−1. ⇒ dy dx = logy+1 1− x y. = y(logy+1) y− y 1+logy [Using (1)] = (1+logy)2 logy. Was this answer helpful? 86. Similar Questions. Q 1.
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Here, we have $g (y)=ay+b$, and therefore, \begin {align}%\label {} \nonumber E [X|Y]=aY+b, \end {align} which is a function of the random variable $Y$. Since $E [X|Y]$ is a random variable, we can find its PMF, CDF, variance, etc. Let's look at an example to better understand $E [X|Y]$.
Covariance formula E [XY ] − E [X ]E [Y ], or “expectation of product minus product of expectations” is frequently useful. Note: if X and Y are independent then Cov(X , Y ) = 0.