# What are the mean and standard deviation of the probability density function given by #p(x)=ke^-x # for # x in [0,1]#, in terms of k, with k being a constant such that the cumulative density across all x is equal to 1?

##### 1 Answer

Mean:

Std Dev:

#### Explanation:

Let

The mean is given by;

# E(X) = int_D \ xp(x) \ dx #

# \ \ \ \ \ \ \ \ \ = int_0^1 xke^(-x) \ dx #

# \ \ \ \ \ \ \ \ \ = k \ int_0^1 xe^(-x) \ dx #

# \ \ \ \ \ \ \ \ \ = k [-(x+1)e^(-x)]_0^1 #

# \ \ \ \ \ \ \ \ \ = -k [(x+1)e^(-x)]_0^1 #

# \ \ \ \ \ \ \ \ \ = -k (2e^(-1)-e^0) #

# \ \ \ \ \ \ \ \ \ = k (1-2/e) #

And, the Variance is given by

# Var(X)=E(X^2)-E^2(X) #

We can calculate,

# E(X^2) = int_D \ x^2p(x) \ dx #

# \ \ \ \ \ \ \ \ \ \ \ = k \ int_0^1 x^2e^(-x) \ dx #

# \ \ \ \ \ \ \ \ \ \ \ = k [-(x^2+2x+2)e^(-x) ]_0^1 #

# \ \ \ \ \ \ \ \ \ \ \ = -k [(x^2+2x+2)e^(-x) ]_0^1 #

# \ \ \ \ \ \ \ \ \ \ \ = -k (5e^(-1) -2e^0]#

# \ \ \ \ \ \ \ \ \ \ \ = k (2-5/e)#

And so the standard deviation is given by

# sigma^2= k (2-5/e) - (k (1-2/e) )^2#

Incidental, As

# int_0^1 \ p(x) \ dx =1 #

# :. int_0^1 \ ke^(-x) \ dx =1 #

# :. [(ke^(-x))/(-1)]_0^1 =1 #

# :. -k [(e^(-x))]_0^1 =1 #

# :. -k (1/e-e^0) =1 #

# :. k (1-1/e) =1 #

# :. k =1/(1-1/e) #

# :. k =e/(e-1) #

And so the probability density function is

# p(x) = e/(e-1) e^(-x) #