Supposewhere x is normal with mean μ and variance σ. Then I see how to derive mode of f ( y ) (distribution of y), as we need to find the value y that makes f ′ ( y ) == 0However, why is mode not simply e μ ? y is a monotonic function of x, and so when x reaches its mode, then y should also reach its mode. The mode of x is its mean ( μ) hence y's mode should be e μ what mistake have I made?

Question

Supposewhere   x is normal with mean   μ and variance   σ. Then I see how to derive mode of   f  (  y  ) (distribution of   y), as we need to find the value y that makes      f    ′    (  y  )  ==  0However, why is mode not simply      e      μ  ?  y is a monotonic function of   x, and so when   x reaches its mode, then y should also reach its mode. The mode of   x is its mean (  μ) hence   y&#039;s mode should be      e      μ  what mistake have I made?

stafninumfu1tf · Accepted Answer

X has lognormal distribution if   X  =      e    Z   where   Z has normal distribution,   Z  ∼  N  (  μ  ,      σ    2    ). However, the density of   X is then given by:  f  (  x  )  =      1          x              2        π                  σ          2                          e          −              1                  2                      σ            2                                                (          ln          ⁡          (          x          )          −          μ          )                2            Differentiating the density with respect to   x we get  −      1                  x        2                    2        π                  σ          2                          e          −              1                  2                      σ            2                                                (          ln          ⁡          (          x          )          −          μ          )                2              −      1          x              2        π                  σ          2                          e          −              1                  2                      σ            2                                                (          ln          ⁡          (          x          )          −          μ          )                2                        ln      ⁡      (      x      )      −      μ              σ      2            1    x  The mode is the value of   x that maximizes the density. Thus, equating the above derivative to zero and simplifying, we get  −  1  −            ln      ⁡      (      x      )      −      μ              σ      2        =  0or  x  =      e          (      μ      −              σ        2            )      To sum up, your mistake is that you have not used the correct density but the equation for the transformation of variables that gives us the random variable with log-normal distribution.

Suppose where x is normal with mean μ and variance σ .

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