Today I wanted to share with you my experience using the services of JustCloud to backup my personal file.

I started looking for a cloud backup solution about 2 years ago, since I had accumulated a considerable amount of personal data I did not want to lose and as I knew that personal backups such as external disks and USB keys are not safe enough (they can get damaged easily, they can get stolen or lost etc…).

I looked at a few different online comparisons  and found out that JustCloud was the best rated one, so I decided I would give it a go.

Their services looks relatively cheap at first sight, but you will very limited with the basic package and very quickly harassed to buy extra options. Sadly, the service focuses on how many devices you use as opposed to how much data you’re using, so it becomes quite expensive when you have multiple computers, which is my case.

All this worked fine for the 2 years I’ve been using their product and lately I received an e-mail saying I had to renew my subscription. I then decided I would try to retrieve some of the data I had previously stored in order to see if it was really worth it compared to a simple Google Drive solution.

I tried both via the UI and via the FTP method, none worked on my first laptop. It was impossible to connect to the FTP and their software was stopping the recovery procedure a few seconds after it started.

I got in touch with their support team via the online chat. As expected, they were completely useless and essentially only asked very basic questions trying to understand my problem… which was “I cannot retrieve my files at all, I want them all back”. After a couple of minutes, the guy told me they were “aware of the issue I was facing” but he needed to escalate my request to the development team who would get back to me about 4 days later.

First, this is completely unacceptable. If you pay for a backup solution the whole point is to have your backup available on demand, not to have to wait days until somebody helps you out. In particular, they should be going absolutely crazy knowing a client cannot recover his files as the is the key of their services; this is in fact the only thing I am paying for!

Needless to say they never got back to me and yesterday suspended my account because I had not renewed. I did get in touch with their subscription department telling them that a technical issue was ongoing on their side and that they should not suspend the account, but never got a proper answer. There is probably nobody behind this department as most answers I got were automatic reminders, signed by a person I could not find on LinkedIn, for example.

The funniest thing is that today I got an e-mail from the same fake sales person, saying that now that my account was suspended, they were offering me a discount to renew, without of course mentioning anything about my issue not being solved.

I did manage to recover part of the files using another computer. But since I had not lost the originals, I compared and there were several files missing! Clearly this service cannot be trusted.

I would strongly encourage you not to trust JustCloud and to advise anybody using their service to recover their data as soon as possible and seek for a new solution.

I have not considered asking for a refund of the past 2 years of services, although I sincerely feel the service I paid for was not provided as my data could not be recovered. If you have experienced the same and wish to push for a refund, get in touch with me and we might be able to get something back if there are more of us.

I have found multiple articles such as this one where other users share the same experience.

Today, I would like to share a little exercise I did to compute the Expected Shortfall of a normal variable. For those of you who are not familiar with this risk measure, it evaluates the average of the $(1-\alpha)$-worst outcomes of a probability distribution (example and formal definition follow).

If you have enough data, the expected shortfall can be empirically estimated. Let’s say we want to compute the expected shortfall at 95% (denoted $ES_\text{95%}$) and that we have 1000 points. The first thing to do is to sort the 1000 points. Then, we take the $1-95\%=5\%$ worst outcomes, which is for 1000 points $1000 \cdot 5\% = 50$ points. Finally, we simply average these points to get our estimated expected shortfall. Nevertheless, this is only an estimate and it will be precise only if you have enough points available.

Sometimes, you can assume the distribution of the points and when these points are a set of returns, it is quite common to assume that they follow some normal distribution $\mathcal{N}(\mu,\sigma^2)$. In this case, you can find a closed form formula which gives you the exact solution for a given $\alpha$, given the parameters $\mu$ and $\sigma$. This result is available in several papers but usually comes without the proof, which I will provide below.

Let’s assume we have a random variable $X \sim \mathcal{N}(\mu,\sigma^2)$. The first thing to do is to re-express this variable in term of a standard normal variable $Z \sim \mathcal{N}(0,1)$. This is easily done as follows:

$$ X = \mu + \sigma Z $$

The next thing to do is to express the Value at Risk at a certain level $\alpha$, which is defined as follows:

$$\text{VaR}_\alpha (X) = \{ Y ~ | ~ \mathbb{P}(X \leq Y) = 1-\alpha \}$$

This can be easily found as well, since:
$$
\begin{align}
\mathbb{P}(X \leq Y) &= \mathbb{P}( \mu + \sigma Z \leq Y)\\
&= \mathbb{P} \left(Z \leq \frac{Y-\mu}{\sigma} \right)\\
&= \Phi \left( \frac{Y-\mu}{\sigma} \right) = 1-\alpha\\
\end{align}
$$
where $\Phi(\cdot)$ is the cumulative standard normal distribution.

We can then deduce that:
$$\text{VaR}_\alpha (X) = Y = \Phi^{-1}(1-\alpha) \sigma + \mu$$
where $\Phi^{-1}(\cdot)$ is the inverse cumulative standard normal distribution and which can be looked up online.

Now, we can actually start working on the closed-form. Let’s first express the expected shortfall in terms of the value at risk:
$$ES_\alpha(X) = \frac{1}{1-\alpha} \int_\alpha^1 \text{VaR}_u(X) du$$

Using our previous result, we can rewrite the definition above as follows:
$$
\begin{align}
ES_\alpha(X) &= \frac{1}{1-\alpha} \int_\alpha^1 ( \Phi^{-1}(1-u) \sigma + \mu ) du\\
&= \frac{1}{1-\alpha} \left( \int_\alpha^1 \Phi^{-1}(1-u) \sigma du + \int_\alpha^1 \mu du \right)\\
&= \frac{1}{1-\alpha} \int_\alpha^1 \Phi^{-1}(1-u) \sigma du + \mu
\end{align}
$$

As you can see, we managed to split the big integral into two parts, one of which is solved very easily. However, there remains an integral with $\Phi^{-1}(\cdot)$ which is a very complex function so we need to find a way around. To do this, we are going to use a change of variable introducing $u = \Phi(y)$. We have then $y = \Phi^{-1}(u)$ and $du = \phi(y)dy$ where $\phi(y)=\frac{1}{\sqrt{2\pi}} \exp \left( – \frac{y^2}{2} \right)$ is the standard normal density function.

$$
\begin{align}
\int_\alpha^1 \Phi^{-1}(1-u) \sigma du &= \int_{\Phi^{-1} (\alpha)}^{\Phi^{-1}(1)} \Phi^{-1}\left(\underbrace{ 1-\Phi(y)}_{\Phi(-y)} \right) \sigma \phi (y) dy \\
&= \int_{\Phi^{-1} (\alpha)}^\infty -y \sigma \phi (y) dy \\
&= \int_{\Phi^{-1} (\alpha)}^\infty -\frac{y \sigma}{\sqrt{2\pi}} \exp \left( – \frac{y^2}{2} \right) dy \\
\end{align}
$$

So, we got rid of $\Phi^{-1}(\cdot)$ and the remaining integral is relatively easy to solve:

$$
\begin{align}
\int_{\Phi^{-1} (\alpha)}^\infty -\frac{y \sigma}{\sqrt{2\pi}} \exp \left( – \frac{y^2}{2} \right) dy &= \frac{\sigma}{\sqrt{2\pi}} \int_{\Phi^{-1} (\alpha)}^\infty -y \exp \left( – \frac{y^2}{2} \right) dy\\
&= \frac{\sigma}{\sqrt{2\pi}} \left( \exp \left( – \frac{y^2}{2} \right)\right)|_{\Phi^{-1} (\alpha)}^\infty\\
&= \frac{\sigma}{\sqrt{2\pi}} \left( 0 – \exp \left( – \frac{\Phi^{-1} (\alpha)^2}{2} \right)\right)\\
&= -\sigma \frac{1}{\sqrt{2\pi}} \exp \left( – \frac{\Phi^{-1} (\alpha)^2}{2} \right)\\
&= -\sigma \phi \left( \Phi^{-1} (\alpha) \right)\\
\end{align}
$$

Now, we just have to include this in our previous result to get:

$$
\begin{align}
ES_\alpha(X) &= \frac{1}{1-\alpha} \int_\alpha^1 \Phi^{-1}(1-u) \sigma du + \mu\\
&= \frac{1}{1-\alpha} \left[ -\sigma \phi \left( \Phi^{-1} (\alpha) \right) \right] + \mu\\
&= \mu – \sigma \cdot \frac{\phi \left( \Phi^{-1} (\alpha) \right)}{1-\alpha}
\end{align}
$$

This is a very good result, since we know that $\lambda( \alpha ) = \frac{\phi \left( \Phi^{-1} (\alpha) \right)}{1-\alpha}$ is a function of $\alpha$ that can be computed very easily, for example in Excel. Below are the results for $\alpha$ from 80% to 99%.

That’s it, with this simple formula, you can compute the expected shortfall of a normally distributed variable without needing any data point.