Undergrad How do I find the expected value and median of a probability density function?

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To find the expected value and median of the probability density function f(x) = 1/2e^-x/2 for x > 0, one must first ensure that the function is normalized, meaning its integral equals 1. The expected value for a continuous random variable is calculated using the integral of x multiplied by the probability density function over its range. The median is the value that divides the probability distribution into two equal halves, typically found by solving the cumulative distribution function for 0.5. Understanding these definitions is crucial for accurately determining the expected value and median. Proper normalization and integration techniques are essential in this process.
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Hey everyone, I have been struggling to find the expected value and median of f(x) = 1/2e^-x/2, for x greater than 0. I am just wondering how I do so? Thank you.
 
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What are the definitions of the expectation and the median for a continuous random variable? :smile:
 
If f(x) is supposed to be a probability density, then it has to be normalized so its integral = 1.
 

Thread 'Proving that convexity implies second order derivative being positive'

There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...
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