1 of 5 Continuous Probability Distributions Objectives By the end of this lesson you should be able to do the following 1 2 3 4 State the properties of a probability density function continuous pmf State the properties of a continuous cdf Calculate probabilities of events in continuous situations Apply this general thinking to the special case of the Normal Distribution Function Review We looked at discrete probability mass functions Aside from a complicated notation they really boil down to addition and subtraction Underlying all the calculations was the idea that probabilities of the discrete sort can be related to an area calculation The critical next step in going to a continuous situation is comparable to moving from a Riemann Sum to a definite integral 1 1 1 1 Example The function f x x for x 1 2 3 creates the geometric sequence S 2 4 8 2 1 This is a discrete infinite sequence 2 You might recall that the sum of this infinite series has a limit of 1 3 Further all values of x are bounded 0 f x 1 Let s use it by saying that for the random variable X S the probability for any whole number x is given by f X x 1 Now we have a pmf 2x We can easily answer questions like this set 1 Calculate the probability that x 5 f X 5 15 1 32 2 1 1 1 21 4 6 2 64 2 2 2 x 1 3 Write the associated cumulative distribution function F X x k k 1 2 4 Calculate the probability that X 10 This equate to F X 10 This might seem like it ll be 2 Calculate the probability that x is 2 4 or 6 f X 2 4 6 tough However the sum for the first n terms of finite geometric sequences is calculated by Sn a0 1 r n 1 r In our case F X x S x 1 2 1 1 1 x 2 1 12 From there it s short jump to F X 10 1 2010 Arizona State University School of Mathematical and Statistical Sciences 1 x 2 12 10 1023 1024 2 of 5 Continuous Probability Distributions Example Look at f X x kx x 0 1 2 3 Find the k that makes this a pmf 4 4 At first look this seems difficult Actually it is quite easy All we need to do is kx k x 1 find the value of K that allows i 1 i 1 4 4 k 0 1 2 3 1 f X x kx 1 for x 0 1 2 3 6k 1 i 1 i 1 We also can save some effort be realizing that the K value is a constant so it can k 16 be factored out of the summation All that work is to the right 2 Example Look at f X x kx x 1 2 3 4 5 4 4 2 Find the k that makes this a pmf kx k x 2 1 You might panic a little but this is almost a clone of the i 1 i 1 previous problem The work is to the right If you wanted to 2 2 2 2 2 you could have used the sum of a quadratic sequence formula k 1 2 3 4 5 1 n n 1 2n 1 6 5 5 1 2 5 1 6 55k 1 k 1 55 55 6 k Example Look at f X x 2 x 1 2 4 8 16 x Find the k that makes this a pmf By now you re getting a little bored with this process This one might spice up the process Please notice you cannot use the quadratic sum formula 2010 Arizona State University School of Mathematical and Statistical Sciences 5 5 k 1 x2 k x2 1 i 1 i 1 1 1 1 1 1 k 2 2 2 2 2 1 4 8 16 1 2 341 k 1 256 k 256 341 3 of 5 Continuous Probability Distributions Continuous Random Variables Let s adopt a subtle change used in some textbooks They refer to the continuous pmf as the Probability Density Function pdf After that there isn t much to change We use the different names just to simply our discussion We need the pdf f x to have the follwing properties to be consistent with the finite situation of the pmf Let S0 be the outcome space of a continuous process For S x X s x s S0 f x is a Probability Density Function when the following are all true 1 0 f x x S 2 S f x dx 1 3 For an event E S0 p E S0 E f x dx The cdf for this function is F x X X a f x dx for x a X Hence the cdf provides the probability for all values below the x value stated X f x dx We could equally write p a x X a The good news Almost any continuous positive valued function with a finite integral on an interval can be made into a pdf We can get there be two routes The first is to find a multiplier coefficient that adjust the value of the integral of the function to 1 The second is to restrict the interval of integration Example Look at f X x kx x 0 1 Find the k that makes this a pdf First don t believe everything you read Confirm that 0 f x x 0 1 It is obvious but do it anyway The real trick is find the k through integration Just solve the equation k The integration gives us k 1 0 xdx k 2 1 0 xdx 1 1 x 2 0 k 1 So k 2 2 It only takes a little more confirmation to see that f X x 2 x x 0 1 is a valid pdf If the question were just find the CDF for x X the answer would be the function not the number Then we X f x dx for X 1 would just answer F X x 0 2010 Arizona State University School of Mathematical and Statistical Sciences 4 of 5 Continuous Probability Distributions Example Look at f X x k x 0 5 1 x Find the k that makes this a pdf If it is not possible say so It is obvious that 0 f x x 0 5 1 when k is positive We still want to think about it Now find the k where F X x 1 0 5 f x dx 1 From the work to right the necessary k is ln 2 If the various manipulations of the logs bothers you review the rules of logarithms now Example Look at f X x 1 0 5 f x dx 1 1 k 0 5 x dx 1 k ln x x 1 x 0 5 1 k 0 ln 0 5 1 1 k ln 2 ln 0 5 k …