$y_2$ is log normal with parameters $(\mu_2, \sigma_2^2)$, then the product $y_1 y_2$ is log normal

with parameters $(\mu_1 + \mu_2, \sigma_1^2 + \sigma_2^2)$.

**Note:** While the product of two log normal distributions is log normal, the **sum** of two log normal distributions is **not** log normal.

```{note}

While the product of two log normal distributions is log normal, the **sum** of two log normal distributions is **not** log normal.

```

This observation sets the stage for challenge that confronts us in this lecture, namely, to approximate probability distributions of **sums** of independent log normal random variables.

<!-- #region -->

Now we shall set a grid length $I$ and a grid increment size $m =1$ for our discretizations.

**Note**: We set $I$ equal to a power of two because we want to be free to use a Fast Fourier Transform

```{note}

We set $I$ equal to a power of two because we want to be free to use a Fast Fourier Transform

to compute a convolution of two sequences (discrete distributions).

```

We recommend experimenting with different values of the power $p$ of 2.

plt.figure(figsize = (8,8))

plt.subplot(2,1,1)

plt.plot(x[:np.int(NT)],p1[:np.int(NT)],label = '')

plt.plot(x[:int(NT)],p1[:int(NT)],label = '')

plt.xlim(0,2500)

count, bins, ignored = plt.hist(s1, 1000, density=True, align='mid')

plt.figure(figsize = (8,8))

plt.subplot(2,1,1)

plt.plot(x[:np.int(NT)],c1f[:np.int(NT)]/m,label = '')

plt.plot(x[:int(NT)],c1f[:int(NT)]/m,label = '')

plt.xlim(0,5000)

count, bins, ignored = plt.hist(ssum2, 1000, density=True, align='mid')

NT= np.size(x)

plt.figure(figsize = (8,8))

plt.subplot(2,1,1)

plt.plot(x[:np.int(NT)],c2f[:np.int(NT)]/m,label = '')

plt.plot(x[:int(NT)],c2f[:int(NT)]/m,label = '')

plt.xlim(0,5000)

count, bins, ignored = plt.hist(ssum3, 1000, density=True, align='mid')

mu7, sigma7 = -.040, 1.4632

```

**Note:** Because the failure rates are all very small, log normal distributions with the

```{note}

Because the failure rates are all very small, log normal distributions with the

above parameter values actually describe $P(A_i)$ times $10^{-09}$.

```

So the probabilities that we'll put on the $x$ axis of the probability mass function and associated cumulative distribution function should be multiplied by $10^{-09}$

```{code-cell} python3

d13 = np.cumsum(c13)

Nx=np.int(1400)

Nx=int(1400)

plt.figure()

plt.plot(x[0:np.int(Nx/m)],d13[0:np.int(Nx/m)]) # show Yad this -- I multiplied by m -- step size

plt.plot(x[0:int(Nx/m)],d13[0:int(Nx/m)]) # show Yad this -- I multiplied by m -- step size

plt.hlines(0.5,min(x),Nx,linestyles='dotted',colors = {'black'})

plt.hlines(0.9,min(x),Nx,linestyles='dotted',colors = {'black'})

plt.hlines(0.95,min(x),Nx,linestyles='dotted',colors = {'black'})