Suppose that X and Y are independent and Z=X+Y. If the probability density functions of X and Y are fx (2) = { if 0
Posted: Wed May 04, 2022 11:57 am
Suppose that X and Y are independent and Z=X+Y. If the probability density functions of X and Y are fx (2) = { if 0 <y <2 if x > 0 otherwise fr (3) = {3, otherwise 0, Which of the following one is the p.d.f. of X+Y? Hint: only one correct answer; If you believe that some math formals are not properly displayed, please switch the browser, say Google Chrome. e-¹(e²-1), ift > 2 fx+y (t) fx+y (t)= fx+y (t) = fx+y (t) fx+y (t): = | // (1 −e-¹), fx+y (t): = Je-¹(e²-1), ift > 2 (1-e¹), fx+y (t) = Je (e4-1), ift > 4 1/(1-e-t), [(e²-1), fx+y (t): (1-et), = = 1/(1-e-¹), Jet(e4-1), ift > 4 (1-e-t), e-(4-1), (1-e¹), Jet(e¹1), (1-e-¹), et(e¹1), if 0 < t < 2 if 0 < t < 4 ift> 4 if 0 < t < 4 ift > 1 if 0 < t < 1 ift > 1 if 0 < t < 1 if 0 < t < 2 if 0 < t < 4 ift > 2 if0<t < 2
Posted: Wed May 04, 2022 11:57 am
Suppose that X and Y are independent and Z=X+Y. If the probability density functions of X and Y are fx (2) = { if 0 <y <2 if x > 0 otherwise fr (3) = {3, otherwise 0, Which of the following one is the p.d.f. of X+Y? Hint: only one correct answer; If you believe that some math formals are not properly displayed, please switch the browser, say Google Chrome. e-¹(e²-1), ift > 2 fx+y (t) fx+y (t)= fx+y (t) = fx+y (t) fx+y (t): = | // (1 −e-¹), fx+y (t): = Je-¹(e²-1), ift > 2 (1-e¹), fx+y (t) = Je (e4-1), ift > 4 1/(1-e-t), [(e²-1), fx+y (t): (1-et), = = 1/(1-e-¹), Jet(e4-1), ift > 4 (1-e-t), e-(4-1), (1-e¹), Jet(e¹1), (1-e-¹), et(e¹1), if 0 < t < 2 if 0 < t < 4 ift> 4 if 0 < t < 4 ift > 1 if 0 < t < 1 ift > 1 if 0 < t < 1 if 0 < t < 2 if 0 < t < 4 ift > 2 if0<t < 2