Visually Identifying Mean of a Probability Density Function
Date: 03/06/2007 at 20:00:24 From: Raghu Subject: Mean, median and mode of a pdf Hi. When I look at a PDF (f(x)), how can I interpret the mean of the PDF? Median is the value of x such that the area of curve to the left of x is equal to the area to the right. Mode is the value of x for which the curve achieves its peak. But, I am not able to visualize what the mean represents in terms of the curve. I understand that the mean is given by the equation sum(x f(x)). But, I want to understand how it is represented (or hidden) in the graph. How can I gauge a value for it by just looking at the graph?
Date: 03/07/2007 at 21:17:06 From: Doctor Wilko Subject: Re: Mean, median and mode of a pdf Hi Raghu, Thanks for writing to Ask Dr. Math! Let's back up for a minute, then come back to your question. Mean, median, and mode are all called measures of "center". Consider a set of data, 1, 2, 2, 2, 3, 6, 7, 8, 23 MODE of a data set is a value that occurs most frequently. It'd be the highest point if you made a histogram of the data. MEDIAN of a data set is the middle value when original data values are arranged in order of increasing (or decreasing) magnitude. Half the values are above, half are below. MEAN (arithmetic mean) of a set of values is the number obtained by adding the values and dividing by total number of values. Using our nine data values from above, we get Mode = 2 (Occurs three times) Median = 3 (Center value, four numbers above and below 3) Mean = 6 (54/9 = 6; Balance point) 2 2 1 2 3 6 7 8 23 -------------------------------------------- /\ The mean is the point where the distribution would balance. Now, let's think about these terms as applied to a probability density function (pdf). Try to see the similarities/connections of these terms as applied to a discrete data set (like above) to a pdf (like below). Take pdf f(x) = 2x; 0 <= x <= 1 Graphically, /| (1,2) / | / | / | / | / | / | / | ---------- | | 0 1 The MODE of a pdf is its highest point, or peak, in this case x = 1. The MEDIAN is the midpoint of probability. It's the point on the x-axis that gives me equal area on both sides. In this case it's x = sqrt(2)/2 ~= 0.7071. Recall, b / P(a <= x <= b) = | f(x) dx / a sqrt(2)/2 / = | 2x dx = 0.50 / 0 The MEAN is the point on the horizontal axis where the distribution would balance. The mean, referred to as the expected value, is defined as (for continuous pdfs) oo / E(x) = | x*f(x) dx = / -oo oo / | x*2x dx = 2/3 ~= 0.6667 / -oo The mean (expected value) is 2/3. 4/9 or about 44% of the area lies to the left of the mean and about 56% of the area lies to the right of the mean. I guess the way to think about it is, if this were a physical object that you'd wish to balance, you'd balance the pdf at x = 2/3. In a symmetrical pdf like the normal distribution, mean = median = mode, but this is not true of skewed distributions. Does this help? Please write back if you have further questions on this. - Doctor Wilko, The Math Forum http://mathforum.org/dr.math/
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