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Lecture 20 Agenda 1. Uniform Distribution Since we have setup a general framework for continuous random variables, let’s look at particular examples. These particular examples are not being studied because they are mathematically pleasing, though that’s one reason; these examples are of immense practical value and we use them everyday. Let’s start with the easiest of them. Uniform Distribution Consider an experiment which consists of choosing a point from the interval [a, b] such that “all points are equally likely to be chosen”. Our random variable X is the point chosen. So if x1 and x2 are two points in [a, b] we are not saying that P (X = x1 ) and P (X = x2 ) are equal, because obviously they are both 0 and everybody knows that. We know that if the pdf at some point x is high, then x lies in a high probability region and if pdf is small then x lies in a low probability region. So here we mean that, inside [a, b] there is no high or low probability region. So the pdf should also be the same everywhere inside [a, b]. In other words, if x ∈ [a, b] otherwise fX (x) = c = 0 where c is some unknown constant. How do we find c ? Well, since fX is a pdf, Z ∞ fX (x)dx = 1 −∞ Z ⇒ b c dx = 1 a ⇒ c = 1 1 b−a In conclusion the density of X is 1 fX (x) = if x ∈ [a, b] b−a = 0 otherwise This density is known as the “UNIFORM DENSITY ON [a, b]” and X is known as a “UNIFORM RANDOM VARIABLE on [a, b]”. It is written as X ∼ U [a, b]. Now, let’s derive some properties of X. Distribution Function FX (x) = 0 Z = x a = 1 if x < a 1 x−a dy = b−a b−a if x > b if a ≤ x ≤ b Hence for a < c < d < b, d−a c−a d−c − = b−a b−a b−a Notice that this depends on length of the interval [c, d] only. This property is a characteristic of the uniform distribution. Caution however this result holds only if [c, d] ⊂ [a, b]. P (X ∈ [c, d]) = FX (d) − FX (c) = Mean and Variance Z ∞ E(X) = xfX (x)dx −∞ Z b = = = = = 1 dx b−a a Z b 1 xdx b−a a 1 b 2 − a2 b−a 2 1 (b + a)(b − a) b−a 2 (b + a) 2 x 2 This result is intuitive. Since all points in [a, b] are equally likely, the mean is the middle of the interval. Let’s find the variance, V (X) = E[(X − E(X))2 ] 2 Z b a+b 1 = x− dx 2 b−a a Now put z = x− a+b 2 b−a hence dx = (b − a)dz, V (X) = (b − a) 2 Z 1 2 z 2 dz − 12 = (b − a) 2 z3 3 12 − 21 (b − a)2 12 = Example Every 15 minutes the bus of at a bus stop let X be the X ∼ U [0, 15]. What’s the probability that Find the time that there is come by that time. route 17 leaves RTS Downtown. If I arrive number of minutes I have to wait. a bus will come in 5 mins ? only 10% chance that the bus will not 5−0 15 − 0 1 = 3 P (X < 5) = For the second question, if a is the answer then, P (X > a) ⇒ P (X ≤ a) a ⇒ 15 ⇒ a = 0.1 = 0.9 = 0.9 = 15 ∗ 0.9 = 13.5 3 Homework :: 1. Let X ∼ U nif orm[a, b] and for any two constants m > 0 and c define Y = mX + c. Prove Y ∼ U nif orm(ma + c, mb + c). 2. Let X be a discrete random variable with the following distribution. X 1.3 2.1 5.7 P (X = x) 0.25 0.3 0.2 8.9 0.25 Now suppose U ∼ U nif orm(0, 1), and based on U define Y as follows if 0 ≤ U < 0.25 if 0.25 ≤ U < 0.55 if 0.55 ≤ U < 0.75 if 0.75 ≤ U < 1 Y = 1.3 = 2.1 = 5.7 = 8.9 Prove that Y has same distribution as X. 3. 4.43, 4.50 4