Question

Suppose we wish to generate a sample from the exponential (β) distribution, and only have access to a computer which generates numbers from the skew logistic distribution. It turns out that if X~SkewLogistic (β), then log(1+exp(-X)) is exponential (β). Show that this is true and check by simulation that this transformation is correct.

Answer 1

The CDF of Skewed logistic distribution is given by .

F(x) = ( 1 + e^-x)^-

Let   F(x) = y

y = ( 1 + e^-x)^- = 1 / ( 1 + e^-x)

( 1 + e^-x) = 1 / y

( 1 + e^-x)   = ( 1 / y )^(1/)

e^-x = ( 1 / y )^(1/) -1

Talking log both side

-x = log( ( 1 / y )^(1/) - 1 )

x = - log( ( 1 / y )^(1/) - 1 )                            ......( 1 ) Skewed logistic distribution

Here y ~ U ( 0 , 1 )

So we will generate random samples for y from Uniform ( 0,1 ) for for some fixed values of

Its transformation i.e X ~ Skew Logistic (β) , then log(1+exp(-X))

Here

x = - log( ( 1 / y )^(1/) - 1 )   

Transforming variable to log(1+exp(-X))

exp( -x ) = ( 1 / y )^(1/) - 1

1 + exp( -x ) = ( 1 / y )^(1/)

log( 1 + exp( -x ) ) = log ( ( 1 / y )^(1/) )

Now CDF of Exponential distribution is given by

F(x) = 1 - e ^{- ( x)}

Let y = F(x)

y = 1 - e ^{- ( x)}

    e ^{- ( x)}   = 1 - y

Talking log both side

- x   = log( 1-y )

x = - log ( 1- y ) /                ......( 2 ) for exponential distribution

Here y ~ U ( 0 , 1 )

{ Note -To generate random variable we can use statistical table or any software like R , Mini-tab etc

   Here I am using R software to generate random variable }

We will generate Random sample from U ( 0 , 1 ) , so that we will get Random samples for Skewed logistic distribution by given in eq (1) and then for exponential distribution by given in eq (2)

R - code

# for Skewed logistic distribution

# let us fix = 2

>y=runif(100,0,1)              # will generate 100 Random sample from U ( 0 , 1 )
>b=2                                 # b = , fix = 2

   # Now " x1 " will generate simulated Random samples for Skewed logistic distribution
>x1 = - log( ( 1 / y )^(1/b) - 1)         

# " TX " is transformation used

>TX=log(1+exp(-x1))                         # given transformation for Skewed logistic

( Note- Histogram and density curve for transformed variable   )

> hist(TX, prob=TRUE,main="Transformed Variable ") # to generate histogram for = 2
> lines(density(TX),col=10)              # to dram density curves

Thus we can see that transformed variable look like exponential distribution .

Run similar program for b = = 5 and draw histogram and curve for it

> hist(TX, prob=TRUE,main="Transformed Variable ") # to generate histogram for = 5
> lines(density(TX),col=10)              # to dram density curves

Thus we can see that transformed variable look like exponential distribution .

Now we will draw random sample for exponential distribution

Note - In R we can use direct command " rexp ( n , b ) " to draw random samples from exponential distribution, where n is sample size { for this case n =100 } and b = is parameter value { this case = 2 , 5 }

But we will use eqⁿ ( 2 ) to generate random sample

>y=runif(100,0,1)                          )              # will generate 100 Random sample from U ( 0 , 1 )
>b=2                                 # b = , fix = 2

   # Now " x3 " will generate Random samples from exponential distribution

>x3=-log(1-y)/b

( Note- Histogram and density curve for generated random variables )
>hist(x3, prob=TRUE,,main="Exponential distribution")           # to generate histogram    for = 2
>lines(density(x3),,col=10)    # to dram density curves

Run similar program for b = = 5 and draw histogram and curve for it

> hist(x3, prob=TRUE,main="Exponential distribution") # to generate histogram for = 5
> lines(density(x3),col=10)              # to dram density curves

Hence we can see If if X~SkewLogistic (β), then log(1+exp(-X)) is exponential (β).

{ In R code ; x1 represents simulated random variables from SkewLogistic (β),

            transformation log(1+exp(-X)) is stored in variable TX and exponential (β). is stored in variable x3 }