Problem 5.2 (Video 3.1, 3.2, 3.3, 3.4, Quick Calculations) Calculate each of the requested quantities. All of these problems are carefully chosen so that they can be completed without integration, so we are expecting exact answers. For the Gaussian problems, you will sometimes need to lookup values for the standard nor- mal CDF @(z). For example, you could use a lookup table https://en.wikipedia.org/wiki/ Standard_normal_table, Wolfram Alpha https://www.wolframalpha.com with query normal cdf calculator, or, in Python, import scipy.stats as st followed by st.norm.cdf(z) where you should replace z with your value. 1] || 1/4]. 0 otherwise (a) For fx(z) = { (b) For fx(z) from part (a), determine and sketch the conditional PDF fx g(z) given the event {|X| > 1/4}. Using your sketch, calculate P[X > 0||X| > 1/4]. (c) Let X be Exponential(2). Calculate E[X 1] and E[(X 1)?%]. (d) Let X be Uniform(2,3). Determine P[X 1 > 0] and P[X2 1 > 0]. (e) Let X be Gaussian(1,4) and let = 3X 2. Determine the mean and variance of Y as well as PY > 1]. (f) Let X be Gaussian(1,4). Determine (up to two decimal places) the maximum value of a such that P[X b] 1/2]. otherwise Solution: Ifx (a) -1 OIN NIH By symmetry, E[X] = 0. For P[|X| > 1/2], we need to sum up the area of two triangles, each with base 1/2 and height 1/2. Therefore, P[|X| > 1/2] = 2 . 2 . 2 . 2 : (b) For fx (x) from part (a), determine and sketch the conditional PDF fx B(x) given the event {IX| > 1/2}. Using your sketch, calculate P[X > 0 | |X| > 1/2]. Solution: fx (2) We have that fx B(2) = { P[X E B] CEB 4 - 4|20 2 0 | |X| > 1/2] - 2 . 2 . 2 = 3. c) Let X be Exponential(2). Calculate E[X - 1] and E[(X - 1)2]. Solution: EX - 1] = E[X] - 1 = 1 2 -1 = E[(X - 1)2] = E[X2 - 2X + 1] = E[X2] -2E[X] + 1 = Var [X] + (E[X])2 -2E[X] + 1 =-+ - 2 2 +1= ? (d) Let X be Uniform(-2, 3). Determine P[X - 1 > 0] and P[X2 - 1 > 0]. Solution: Note that the PDF of X is has height 1/5 from x = -2 to x = 3 and is 0 otherwise. Therefore, all of our calculations can be carried out as areas of rectangles. P[X - 1 > 0] = P[X > 1] = = (3-1) - P[X2 - 1 > 0] = P[X2 > 1] = P[X > 1] + P[X 1]. Solution: By the linearity of expectation, E[Y] = E[3X 2] =3E[X]-2=3-2-2=4. Using the variance of a linear function, Var[Y] = Var[3X 2] = 3*Var[X] =9-1 =0. 1-4 PlY >1]=1-F1)=1 'D(T) 1-(-1) ~0.8413 (f) Let X be a uniform random variable distributed over the interval (0,6). Compute the variance of X. If Z = e~X, compute E[Z]. Solution: Var(X) = % = 3, using the fact that the variance for uniform random variables is the square of the length, divided by 12. 6 E[Z] = E[e X] = /D exdm _ u _e )