Loss function for Logistic Regression

If we are doing a binary classification using logistic regression, we often use the cross entropy function as our loss function. More specifically, suppose we have $$T$$ training examples of the form $$(x^{(t)},y^{(t)})$$, where $$x^{(t)}\in\mathbb{R}^{n+1},y^{(t)}\in\{0,1\}$$, we use the following loss function $$\mathcal{LF}(\theta)=-\dfrac{1}{T}\sum_{t}y^{t}\log(\text{sigm}(\theta^T x))+(1-y^{(t)})\log(1-\text{sigm}(\theta^T x))\,,$$ where $$\text{sigm}$$ denotes the sigmoid function.

Question: However, if we are doing linear regression, we often use squared-error as our loss function. Are there any specific reasons for using the cross entropy function instead of using squared-error or the classification error in logistic regression?

I read somewhere that, if we use squared-error for binary classification, the resulting loss function would be non-convex. Is this the only reason, or is there any other deeper reason which I am missing?

Attempt: To get a sense of what different loss functions would look like, I have generated $$50$$ random data points on both sides of the line $$y=x$$. I have assigned the class $$c=1$$ to the data points which are present on one side of the line $$y=x$$, and $$c=0$$ to the other data points. After generating this data, I have computed the costs for different lines $$\theta_1 x-\theta_2y=0$$ which pass through the origin using the following loss functions:

1. squared-error function using the predicted labels and the actual labels.
2. squared-error function using the continuous scores $$\theta^Tx$$ instead of thresholding by $$0$$.
3. squared-error function using the continuous scores $$\text{sigm}(\theta^T x)$$.
4. classification error, i.e., number of misclassified points.
5. cross entropy loss function.

I have considered only the lines which pass through the origin instead of general lines, such as $$\theta_1x-\theta_2y+\theta_0=0$$, so that I can plot the loss function. I have obtained the following plots. From the above plots, we can infer the following:

1. The plot corresponding to $$1$$ is neither smooth, it is not even continuous, nor convex. This makes sense since the cost can take only finite number of values for any $$\theta_1,\theta_2$$.
2. The plot corresponding to $$2$$ is smooth as well as convex.
3. The plot corresponding to $$3$$ is smooth but is not convex.
4. The plot corresponding to $$4$$ is neither smooth nor convex, similar to $$1$$.
5. The plot corresponding to $$5$$ is smooth as well as convex, similar to $$2$$.

If I am not mistaken, for the purpose of minimizing the loss function, the loss functions corresponding to $$(2)$$ and $$(5)$$ are equally good since they both are smooth and convex functions.

Is there any reason to use $$(5)$$ rather than $$(2)$$? Also, apart from the smoothness or convexity, are there any reasons for preferring cross entropy loss function instead of squared-error?

You got off on the wrong track as detailed here. Just because you have a binary $Y$ it doesn't mean that you should be interested in classification. You are really interested in a probability model, so logistic regression is a good choice. Get the nomenclature right or you will confuse everyone.