Question

Using the inverse transform method...

4.2 Inverse-Transform Method 2, where l < t < 5, Explain how to generate values from a continuous distribution with density function/() = given u E O,1).

Answer 1

QUestion 4.2

As we know that probability distribution have probabilities between 0 to 1.

So similarly, cumuative probability distribution is also lies between 0 to 1.

so Here

f(t) = 5/4t^-2   ; 1 < t < 5

so cumulative probability distribution .

F(t) = $\int_{1}^{t} f(t)dt$    = $\int_{1}^{t} 5t^{-2}dt/4$ = -5/4 [t^-1 ]^t₁ = 5/4 (1 - 1/t)

F(t) = 5/4(1- 1/t) ; 1 < t < 5

sso now we know that 0 < F(t) < 1

so if we take

u = 5/4 (1 - 1/t)

4u/5= 1 -1/t

1/t = 1 - 4u/5

1/t = (5 - 4u)/5

t = 5/(5-4u)

So we can get values of t as per the the distribution for any random value generator of values from u ~ (0,1)

I am posting here 25 random numbers

u	t
0.2567	1.2585
0.7595	2.5483
0.4698	1.6021
0.8554	3.1674
0.6384	2.0437
0.0855	1.0734
0.2969	1.3114
0.9349	3.9666
0.7474	2.4869
0.9571	4.2678
0.5395	1.7594
0.4387	1.5407
0.3423	1.3771
0.6993	2.2700
0.9188	3.7745
0.8764	3.3458
0.8252	2.9422
0.8113	2.8491
0.9088	3.6636
0.7530	2.5150
0.6336	2.0279
0.2468	1.2460
0.6667	2.1431
0.5996	1.9219
0.2476	1.2470